500_AN-INTRO DUCTION TO GENETICS_ANALYSIS_711

446 Chapter 13 • The Dynamic Genome: Transposable Elements

unknown mechanism, triggered the suppression of both resistance gene or other introduced transgenes. In addi-
itself and the homologous gene in the petunia chromo- tion, several transposable elements normally inactive in
some. This phenomenon has come to be known as the genome of Clamydomonas were reactivated in the
cosuppression. mutant strains and shown to insert into new chromoso-
mal locations.
As the transformation of some plant species with
foreign genes became routine, scientists noticed that a Similar results in plants and animals have made it
variety of transgenes were efficiently silenced in the increasingly clear that epigenetic regulation serves not
plant host genome. Because the transgenes could often only as an effective way to silence cellular genes (Chap-
be reactivated, this silencing was recognized as a form of ter 10) but also as a major line of defense against the po-
epigenetic regulation. However, it is highly unlikely that tentially mutagenic effects of transposon activity.
organisms have evolved mechanisms to turn off trans-
genes introduced by plant scientists. Instead, it was A genomic battleground?
reasoned that transgenes are silenced because they re-
semble a natural threat to the host, perhaps their own We have already seen that transposable elements have
transposable elements or infecting viruses or both. Like caused a variety of mutations in plants and animals.
transposable elements and viruses, transgenes can insert Thus, there must be times when host regulation of trans-
into new sites in the host genome. What if organisms posable elements can be overcome and silenced ele-
had defense mechanisms that could recognize these “in- ments are reactivated. Looked at another way, if host
vaders” and turn them off by silencing their expression, regulation were completely successful, transposable
possibly through changes in chromatin structure? elements would no longer exist; they would be silenced,
unable to transpose, and would gradually mutate into
In an attempt to identify host genes contributing to unrecognizable sequences. Rather, there appears to be a
transgene and possibly transposable-element silencing, constant battle between the proliferation of transposable
geneticists sought suppressor strains that had lost the elements and host attempts to silence or otherwise inac-
ability to silence transgenes. One approach used a strain tivate them.
of the unicellular green alga Chlamydomonas rheinhardii
that contained a silenced transgene that normally con- In this regard, some of you may now be concerned
ferred resistance to the antibiotic spectinomycin. This that almost 50 percent of your genome is derived from
strain, which could not grow on agar plates containing transposable elements. There is really no need to worry.
spectinomycin, was treated with a mutagen and then Humans and all other organisms have coevolved with
spread onto plates containing the antibiotic. A cell with their transposable elements and have worked out a vari-
a mutation in a gene required to silence the transgene ety of mechanisms so that both are able to coexist. Or-
should be able to grow on these plates. Mutant strains ganisms not able to evolve a satisfactory accommodation
were indeed isolated in this way and, as expected, these with their transposable elements have most likely be-
strains were unable to silence the spectinomycin- come extinct.

KEY QUESTIONS REVISITED either side by IS elements. The IS elements move them-
selves and the gene between them to plasmids that can
• Why were transposable elements first discovered pass to nonresistant cells though bacterial conjugation.
genetically in maize but first isolated molecularly
from E. coli? • Why are transposable elements classified as RNA
transposons or DNA transposons?
The genetic behavior of transposable elements in maize
genes produced striking kernel phenotypes that were de- RNA transposons, also called class 1, include retrotrans-
ciphered by maize geneticists, especially Barbara posons (LINEs and LTR-retrotransposons) and SINEs
McClintock. However, the maize genome is huge (about (such as the human Alu). For all RNA elements, the
the size of the human genome) and the molecular isola- transposition intermediate is RNA. In contrast, the trans-
tion of maize elements was not possible until decades af- position intermediate of all DNA elements, also called
ter their genetic discovery. In contrast, gene isolation was class 2, is DNA.
pioneered in E. coli (>1000-fold smaller genome than
that of maize) where the first elements to be cloned • How do autonomous and nonautonomous
were the IS elements from E. coli mutations. transposable elements differ?

• How do transposable elements participate in the Autonomous elements encode all the proteins necessary
spread of antibiotic-resistant bacteria? to mobilize themselves and the nonautonomous ele-
ments in their family. Nonautonomous elements rely on
Antibiotic-resistance genes are frequently found in the
chromosome or on plasmids, where they are flanked on

Key terms 447

autonomous elements for their movement because they host regulatory mechanisms. Finally, the vast majority of
do not encode the necessary proteins, including reverse transposable-element sequences in the human genome
transcriptase (for RNA elements) and transposase (for are in noncoding DNA including telomeres, centromeres,
DNA elements). intergenic DNA, and introns.

• How can humans survive given that up to 50 percent • How can the study of retrotransposons in yeast lead
of the human genome is derived from transposable to improved procedures for human gene therapy?
elements?
Yeast retrotransposons target their new insertions to so-
Three major reasons. First, most of the transposable called safe havens, regions of the genome with few genes.
elements’ sequences are mutant and no longer capable By understanding the underlying mechanisms, scientists
of transposition. Second, the transposition of the few may be able to devise new strategies to target genes for
active elements in the genome is usually prevented by gene therapy into safe havens in the human genome.

SUMMARY Ac, Ds, and P are examples of DNA transposons, so
named because the transposition intermediate is the
Transposable elements were discovered in maize by DNA element itself. Autonomous elements such as Ac
Barbara McClintock as the cause of several unstable mu- encode a transposase that binds to the ends of au-
tations. Ds is an example of a nonautonomous element tonomous and nonautonomous elements and catalyzes
that requires the presence of the autonomous Ac excision of the element from the donor site and reinser-
element in the genome for it to transpose. tion into a new target site elsewhere in the genome.

Bacterial insertion sequence (IS) elements were the Retrotransposons were first molecularly isolated
first transposable elements isolated molecularly. There from yeast mutants and their resemblance to retro-
are many different types of IS elements in E. coli strains, viruses was immediately apparent. Retrotransposons are
and they are usually present in at least several copies. class 1 elements, as are all transposable elements that use
Composite transposons contain IS elements flanking one RNA as their transposition intermediate.
or more genes, such as genes conferring resistance to an-
tibiotics. Transposons with resistance genes can insert The active transposable elements isolated from such
into plasmids and are then transferred by conjugation to model organisms as yeast, Drosophila, E. coli, and maize
nonresistant bacteria. constitute a very small fraction of all the transposable
elements in the genome. DNA sequencing of whole
There are two major groups of transposable ele- genomes, including the human genome, has led to the
ments in eukaryotes: class 1 retroelements and class 2 remarkable finding that almost half of the human
DNA elements. The P element was the first class 2 DNA genome is derived from transposable elements. Organ-
transposon to be isolated molecularly. It was isolated isms coexist with their elements largely because epige-
from unstable mutations in Drosophila that were in- netic mechanisms have evolved to suppress the move-
duced by hybrid dysgenesis. P elements have been devel- ment of the elements.
oped into vectors for the introduction of foreign DNA
into Drosophila germ cells.

KEY TERMS C-value (p. 440) junk DNA (p. 444)
C-value paradox (p. 440)
Activator (Ac) (p. 425) Dissociation (Ds) (p. 425) long interspersed nuclear element
Alu (p. 441) DNA element (p. 425) (LINE) (p. 440)
autonomous element (p. 427) DNA transposon (p. 436)
change in phase (p. 445) epimutation (p. 445) long terminal repeat
class 1 element (p. 424) excise (p. 426) (LTR) (p. 434)
class 2 element (p. 424) HeT-A (p. 444)
cointegrate (p. 432) hybrid dysgenesis (p. 436) LTR-retrotransposon (p. 435)
composite transposon (p. 431) insertion-sequence (IS)
conservative transposition (p. 432) M cytotype (p. 436)
copia-like element (p. 435) element (p. 429)
cosuppression (p. 446) inverted repeat (IR) (p. 431) negative selection (p. 442)
“cut and paste” (p. 432)
nonautonomous
element (p. 427)

P cytotype (p. 436)

P element (p. 436)

448 Chapter 13 • The Dynamic Genome: Transposable Elements

provirus (p. 434) R plasmid (p. 431) TART (p. 444)
replicative transposition (p. 432) safe haven (p. 442) transposase (p. 430)
retro-element (p. 424) short interspersed nuclear element transpose (p. 426)
retrotransposition (p. 424) transposition (p. 431)
retrotransposon (p. 435) (SINE) (p. 440) transposon (Tn) (p. 431)
retrovirus (p. 434) silenced (p. 445) transposon tagging (p. 439)
reverse transcriptase (p. 432) simple transposon (p. 431) Ty element (p. 434)
R factor (p. 430) synteny (p. 442) unstable phenotype (p. 425)
RNA element (p. 424) targeting (p. 443)
target-site duplication (p. 432)

SOLVED PROBLEMS

1. In Chapter 10, we studied the operon model. Note but not of upstream genes. Therefore, we would expect
that, for the gal operon, the order of transcription of the insertion mutation to prevent the expression of the
the genes in the operon is E-T-K. Suppose we have galK gene. Three mutations are in this category, gal-1,
five different mutations in galT: gal-1, gal-2, gal-3, gal-2, and gal-3. These mutations could be frameshifts,
gal-4, and gal-5. The following table shows the ex- nonsense mutations, or insertions, because each of them
pression of galE and galK in mutants carrying each of can lead to polarity. If we examine the reversion data,
these mutations: however, we can distinguish among these possibilities.
Transposable elements revert at low rates spontaneously,
galT Expression Expression and this rate is not stimulated by base analogs, frameshift
mutation of galE of galK mutagens, alkylating agents, or UV light. On the basis of
these criteria, the gal-3 mutation is most likely to result
gal-1 1 2 from an insertion, because it reverts at a low rate that is
gal-2 1 2 not stimulated by any of the mutagens, gal-1 might be a
gal-3 1 2 frameshift, because it does not revert with 2-AP and
gal-4 1 1 EMS but does revert with ICR191, a frameshift muta-
gal-5 1 1 gen, and UV light. (Refer to Chapter 16 for details of
each mutagen.) Likewise, gal-2 is probably a frameshift,
In addition, the reversion patterns of these muta- because it reverts only with ICR191. The gal-4 mutation
tions with several mutagens that we studied in is probably a deletion, because it is not stimulated to re-
Chapter 14 are shown in the following table. Here, a vert at all. The gal-5 mutation appears to be a base sub-
“1” indicates a high rate of reversion in the presence stitution, because it reverts with 2-AP, but not above the
of each mutagen, a “2” depicts no reversion, and a spontaneous background rate with ICR191.
“low” indicates a low rate of reversion.
2. Transposable elements have been referred to as “jump-
Reversion ing genes” because they appear to jump from one posi-
tion to another, leaving the old locus and appearing at a
Spon- 2-Amino new locus. In light of what we now know concerning
Mutation taneous purine ICR191 UV EMS the mechanism of transposition, how appropriate is the
term “jumping genes” for bacterial transposable elements?
gal-1 2 2 1 12
gal-2 Solution
gal-3 2 2 1 22
gal-4 In bacteria, transposition takes place by two different
gal-5 Low Low Low Low Low modes. The conservative mode results in true jumping
genes, because, in this case, the transposable element ex-
2 2 2 22 cises from its original position and inserts at a new posi-
tion. A second mode is termed the replicative mode. In
Low 1 Low 1 1 this pathway, a transposable element moves to a new lo-
cation by replicating into the target DNA, leaving be-
Which mutation is most likely to result from the in- hind a copy of the transposable element at the original
sertion of a transposable element such as IS1 and site. When operating by the replicative mode, transpos-
why? Can you assign the other mutations to other able elements are not really jumping genes, because a
categories? copy does remain at the original site.

Solution
Transposable elements will cause polarity, preventing the
expression of genes downstream of the point of insertion

Problems 449

PROBLEMS 8. In Drosophila, M. Green found a singed allele (sn)
with some unusual characteristics. Females homozy-
BASIC PROBLEMS gous for this X-linked allele have singed bristles, but
they have numerous patches of sn+ (wild-type) bris-
1. Suppose that you want to determine whether a new tles on their heads, thoraxes, and abdomens. When
mutation in the gal region of E. coli is the result of these flies are mated with sn males, some females
an insertion of DNA. Describe a physical experi- give only singed progeny, but others give both
ment that would allow you to demonstrate the pres- singed and wild-type progeny in variable propor-
ence of an insertion. tions. Explain these results.

2. Explain the difference between the replicative and 9. Consider two maize plants:
the conservative modes of transposition. Briefly de-
scribe an experiment demonstrating each of these a. Genotype C/c m ; Ac/Ac+, where c m is an unstable
modes in prokaryotes. allele caused by Ds insertion

3. Describe the generation of multiple drug-resistance b. Genotype C/c m, where cm is an unstable allele
plasmids. caused by Ac insertion.

4. Briefly describe the experiment that demonstrates What phenotypes would be produced and in what
that the transposition of the Ty1 element in yeast proportions when (1) each plant is crossed with a
takes place through an RNA intermediate. base-pair-substitution mutant c/c and (2) the plant in
part a is crossed with the plant in part b? Assume that
5. Explain how the properties of P elements in Ac and c are unlinked, that the chromosome breakage
Drosophila make gene-transfer experiments possible frequency is negligible, and that mutant c/C is Ac+.
in this organism.
10. You meet your friend, the scientist, at the gym and
6. Nobel prizes are usually awarded many years after she begins telling you about a mouse gene she is
the actual discovery. For example, Watson, Crick, studying in the lab. The product of this gene is an
and Wilkens won the Nobel Prize in Medicine or enzyme required to make the fur brown. The gene is
Physiology in 1962, almost a decade after their dis- called FB and the enzyme is called FB enzyme.
covery of the double helical structure of DNA. When FB is mutant and cannot produce the FB en-
However, Barbara McClintock was awarded her zyme, the fur is white. The scientist tells you that
Nobel Prize in 1983, almost four decades after her she has isolated the gene from two mice with brown
discovery of transposable elements in maize. Why fur and that surprisingly she found that the two
do you think it took this long? genes differ by the presence of a 250 bp SINE (like
the human Alu element) in the FB gene of one
CHALLENGING PROBLEMS mouse but not the gene from the other. She does
not understand how this is possible, especially when
7. Prior to the integration of a transposon, its trans- she determined that both mice make the FB en-
posase makes a staggered cut in the host target zyme. Can you help her formulate a hypothesis that
DNA. If the staggered cut occurs at the sites of the explains why the mouse can still produce FB en-
arrows below, draw what the sequence of the host zyme with a transposable element in its FB gene?
DNA will be after the transposon is inserted. You
can represent the transposon as a rectangle. 11. The yeast genome has class 1 elements (Ty1, Ty2,
etc.) but no class 2 elements. Can you think of a
p possible reason why DNA elements have not been
successful in the yeast genome?
AATTTGGCCTAGTACTAATTGGTTGG

TTAAACCGGATCATGATTAACCAACC
q



44200_14_p451-480 3/18/04 11:01 AM Page 451

14

MUTATION, REPAIR,
AND RECOMBINATION

KEY QUESTIONS

• What is the molecular nature of mutations?
• How do certain types of radiation and

chemicals cause mutation?
• Are induced mutations different from

spontaneous mutations?
• Can a cell repair mutations?
• What is the molecular mechanism of

crossing-over?
• Do mutational repair systems participate

in crossing-over?
OUTLINE
14.1 Point mutations
14.2 Spontaneous mutation
14.3 Biological repair mechanisms
14.4 The mechanism of meiotic crossing-over

Computer model of a Holliday junction. [Julie Newdol, Computer
Graphics Laboratory, University of California, San Francisco.
Copyright by Regents, University of California.]

451

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452 Chapter 14 • Mutation, Repair, and Recombination

CHAPTER OVERVIEW tion, effected by recombination. As its name suggests,
recombination is the outcome of cellular processes that
Genetic variation among individuals provides the raw cause alleles of different genes to become grouped in
material for evolution. Because genetics is the study new combinations. To use an analogy, mutation produces
of inherited differences, genetic analysis would not be new playing cards, and then recombination shuffles
possible without variants — individuals that show phe- them and deals them out as different hands.
notypic differences in one or more particular characters.
In previous chapters we performed many analyses of the In the cellular environment, DNA molecules are not
inheritance of such variants; now we consider their ori- absolutely stable; each base pair in a DNA double helix
gin. How do genetic variants arise? has a certain probability of mutating. As we shall see, the
term mutation covers a broad array of different kinds of
Two major processes are responsible for genetic vari- changes. In the next chapter, we shall consider mutational
ation, mutation and recombination. We have seen that changes that affect entire chromosomes or large pieces of
mutation is a change in the DNA sequence of a gene. chromosomes. In the present chapter, we focus on muta-
Mutation is the ultimate source of evolutionary change; tional events that take place within individual genes. We
new alleles arise in all organisms, some spontaneously, call such events gene mutations. Many kinds of gene alter-
others as a result of exposure to radiation and chemicals ations can occur within DNA molecules. These events can
in the environment. The new alleles produced by muta- be as simple as the swapping of one base pair for another.
tion become the raw material for a second level of varia- Alternatively, some mutations entail a change in the num-

CHAPTER OVERVIEW Figure

DNA alteration, Crossover at meiosis Double-strand break,
mutation, and repair A mutation, and repair

a A
Double-strand
Crossover initiated by break in gene A
double-strand break
Sister
Gene A chromatid
contributes
Mutation on Nonsister chromatid DNA
one strand contributes DNA
Break
Altered strand Resolution forms crossover healed

Removal Key
= parallels
Repair of mismatches = heterozygous site

Resynthesis

Figure 14-1 Parallels between recombination and certain types of mutational repair.

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14.1 Point mutations 453

ber of copies of a trinucleotide repeat sequence [as when point mutations that alter one base pair at a time. Essen-
(AGC)3 becomes (AGC)5]. Mutations can even be caused tially, these “point” mutations are the minimum changes
by the insertion of a transposable element from elsewhere that can be produced — changing only one “letter” in the
in the genome (Chapter 13). In this chapter we focus on “book of DNA.” The constellation of possible ways in
mutations that do not involve transposable elements. which a point mutation could change a wild-type gene is
very large. However, it is always true that such muta-
We can view DNA as being subjected to a dynamic tions are more likely to reduce or eliminate gene func-
tug of war between the chemical processes that damage tion (thus they are loss-of-function mutations) than to
DNA and lead to new mutations and the cellular repair enhance it (gain-of-function mutations). The reason is
processes that constantly monitor DNA for such damage simple: by randomly changing or removing one of the
and correct it. Mutations often arise through the action components of a machine, it is much easier to break it
of certain agents, called mutagens, that increase the rate than to alter the way that it works. Conversely, muta-
at which mutations occur. Alternatively, mutations can tions that increase a gene’s activity or alter the type of
occur “spontaneously.” Spontaneous mutations are much activity or change the location within a multicellular
less frequent (and hence harder to study) than induced body where the gene is expressed are much rarer.
mutations, but they are evolutionarily more important.
A host of different molecular mechanisms underlie mu- The origin of point mutations
tation, ranging from the reaction of DNA with highly re-
active products of cell metabolism to mistakes in the Newly arising mutations are categorized as induced or
DNA replication process. spontaneous. Induced mutations are defined as those that
arise after purposeful treatment with mutagens, environ-
Cells have evolved sophisticated systems to identify mental agents that are known to increase the rate of mu-
and repair damaged DNA, thereby preventing the occur- tations. Spontaneous mutations are those that arise in
rence of mutations. Most notably, there are a variety of re- the absence of known mutagen treatment. They account
pair systems, and most of them rely on DNA complemen- for the “background rate” of mutation and are presum-
tarity. That is, they use one DNA strand as a template for ably the ultimate source of natural genetic variation that
the correction of DNA damage. For example, in the type is seen in populations.
of repair called excision repair, damage in one strand is cut
out along with adjacent nucleotides, and then the correct The frequency at which spontaneous mutations oc-
sequence is resynthesized using the undamaged comple- cur is low, generally in the range of one cell in 105 to
mentary strand as template (Figure 14-1, left column). 108. Therefore, if a large number of mutants are required
for genetic analysis, mutations must be induced. The in-
Finally, we will see that what is potentially the most duction of mutations is accomplished by treating cells
serious class of DNA damage, a double-strand break, is with mutagens. The production of mutations through
also an intermediate step in a normal cellular process, exposure to mutagens is called mutagenesis, and the or-
recombination via meiotic crossing-over. Thus, we can ganism is said to be mutagenized. The most commonly
draw parallels between mutation and recombination at used mutagens are high-energy radiation or specific
two levels. First, as mentioned above, mutation and re- chemicals; examples of these mutagens and their effi-
combination are the major sources of variation. Second, cacy are given in Table 14-1. The greater the dose of
mechanisms of DNA repair and recombination share mutagen, the greater the number of mutations induced,
some features, including the use of some of the same as shown in Figure 14-2. Note that Figure 14-2 shows a
proteins. For this reason, we will explore mechanisms of linear dose response, which is often observed in the in-
DNA recombination and compare these with mecha- duction of point mutations.
nisms of DNA repair. Figure 14-1 diagrams the parallels
between crossing-over and two kinds of mutational re- Recognize that the distinction between induced and
pair (excision and double-strand break repair). spontaneous is purely operational. If we are aware that
an organism was exposed to a mutagen, then we surmise
We consider two general classes of gene mutation: that the bulk of the mutations that arise afterward were
induced by that mutagen. However, this is not true in an
• Mutations affecting single base pairs of DNA absolute sense. The mechanisms that give rise to sponta-
neous mutations are also acting in this mutagenized
• Mutations altering the number of copies of a small organism. In reality, there will always be a subset of mu-
repeated sequence within a gene tations recovered after mutagenesis that arose indepen-
dently of the action of the mutagen. The proportion of
14.1 Point mutations mutations that fall into this subset depends on how po-
tent a mutagen is. The higher the rate of induced muta-
Point mutations typically refer to alterations of single tions, the lower the proportion of recovered mutations
base pairs of DNA or of a small number of adjacent base that are actually “spontaneous” in origin.
pairs — that is, mutations that map to a single location,
or “point,” within a gene. Here we will focus on the

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454 Chapter 14 • Mutation, Repair, and Recombination

Table 14-1 Mutation Frequencies Obtained with Various Mutagens in Neurospora

Mutagenic treatment Exposure time Survival (%) Number of
(minutes) ad-3 mutants per
No treatment (spontaneous rate) 100
Amino purine (1 – 5 mg/ml) – 100 106 survivors
Ethylmethanesulfonate (1%) During growth
Nitrous acid (0.05 M) 56 ~0.4
X rays (2000 r/min) 90 23 3
Methyl methanesulfonate (20 mM) 160 16
UV rays (600 erg/mm2/min) 26 25
Nitrosoguanidine (25 mM) 18 18 128
ICR-170 acridine mustard (5 mg/ml) 300 65 259
28 350
6 375
240 1500
480 2287

Note: The assay measures the frequency of ad-3 mutants. It so happens that such mutants are red, so they can be
detected against a background of white ad-3ϩ colonies.

16Percent mutations Types of point mutations
15
Point mutations are classified in molecular terms in
14 Table 14-2, which shows the main types of DNA
changes and their effects on protein function when they
13 occur within the protein-coding region of a gene.

12 The two main types of point mutation in DNA are
base substitutions and base additions or deletions. Base sub-
11 stitutions are mutations in which one base pair is replaced
10 by another. Base substitutions also can be divided into two
subtypes: transitions and transversions. To describe these
9 subtypes, we consider how a mutation alters the sequence
8 on one DNA strand (the complementary change will take
place on the other strand). A transition is the replacement
7 of a base by the other base of the same chemical category
(purine replaced by purine: either A to G or G to A;
6 pyrimidine replaced by pyrimidine: either C to T or T to
5 C). A transversion is the opposite — the replacement of a
4 base of one chemical category by a base of the other
(pyrimidine replaced by purine: C to A, C to G, T to A, T
3 to G; purine replaced by pyrimidine: A to C, A to T, G to
C, G to T). In describing the same changes at the double-
2 stranded level of DNA, we must represent both members
of a base pair in the same relative location. Thus, an exam-
1 ple of a transition would be G · C : A · T; that of a
transversion would be G · C : T · A.
0 1 2 3 4 56
Dose (krad) Addition or deletion mutations are actually additions
or deletions of nucleotide pairs; nevertheless, the conven-
Figure 14-2 Linear relationship between X-ray dose and tion is to call them base-pair additions or deletions. Col-
mutation. The relationship is measured by the induction of sex- lectively, they are termed indel mutations (for insertion-
linked recessive lethals in Drosophila. deletion). The simplest of these mutations are single-
base-pair additions or single-base-pair deletions. Muta-
Induced and spontaneous mutations arise by generally tions sometimes arise through the simultaneous addition
different mechanisms, and so they will be covered sepa- or deletion of multiple base pairs at once. As we shall see
rately. After considering these mechanisms, we shall explore later in this chapter, mechanisms that selectively produce
the subject of biological mutation repair. Without these certain kinds of multiple-base-pair additions or deletions
repair mechanisms, the rate of mutation would be so high are the cause of certain human genetic diseases.)
that cells would accumulate too many mutations to remain
viable and capable of reproduction. Thus, the mutational
events that do occur are those rare events that have some-
how been overlooked or bypassed by the repair processes.

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14.1 Point mutations 455

Table 14-2 Point Mutations at the Molecular Level

Type of mutation Result and examples
At DNA level

Transition Purine replaced by a different purine, or pyrimidine replaced by a different pyrimidine:
Transversion
Indel A · T !: G · C !: G · C !: A · T C · G !: T · A T · A !: C · G

At protein level Purine replaced by a pyrimidine, or pyrimidine replaced by a purine:
Synonymous mutation
Missense mutation A · T !: C · G A · T !: T · A G · C !: T · A G · C !: C · G
T · A !: G · C T · A !: A · T C · G !: A · T C · G !: G · C
Conservative missense mutation
Addition or deletion of one or more base pairs of DNA (inserted or deleted bases are
Nonconservative missense mutation underlined):

Nonsense mutation AAGACTCCT !: AAGAGCTCCT
AAGACTCCT !: AAACTCCT
Frameshift mutation
Codons specify the same amino acid:

AGG !: CGG

Arg Arg

Codon specifies a different amino acid
Codon specifies chemically similar amino acid:

AAA !: AGA

Lys Arg
(basic) (basic)

Does not alter protein function in many cases

Codon specifies chemically dissimilar amino acid:

UUU !: UCU

Hydrophobic Polar
phenylalanine serine

Codon signals chain termination:

CAG !: UAG

Gln Amber
termination
codon

One base-pair addition (underlined)

AAG ACT CCT !: AAG AGC TCC T...

One base-pair deletion (underlined)

AAG ACT CCT !: AAA CTC CT...

The molecular consequences of point • Synonymous mutations. The mutation changes one
mutations on gene structure and expression codon for an amino acid into another codon for that
same amino acid. Synonymous mutations are also
What are the functional consequences of these differ- referred to as silent mutations.
ent types of point mutations? First, consider what hap-
pens when a mutation arises in a polypeptide-coding • Missense mutations. The codon for one amino acid
part of a gene. For single-base substitutions, there are is changed into a codon for another amino acid.
several possible outcomes, but all are direct conse- Missense mutations are sometimes called non-
quences of two aspects of the genetic code: degeneracy synonymous mutations.
of the code and the existence of translation termination
codons (Figure 14-3). • Nonsense mutations. The codon for one amino acid is
changed into a translation termination (stop) codon.

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456 Chapter 14 • Mutation, Repair, and Recombination

Thr Gln Arg Gly Wild-type gene tide sequence that extend far beyond the site of the mu-
Codon 1 Codon 2 Codon 3 Codon 4 tation itself (see Figure 14-3). Recall that the sequence
ACACAGCGTGGT of mRNA is “read” by the translational apparatus in
register (“in frame”), three bases (one codon) at a time.
Thr Gln Arg Gly The addition or deletion of a single base pair of DNA
changes the reading frame for the remainder of the
ACACAGCGCGGT translation process, from the site of the base-pair muta-
tion to the next stop codon in the new reading frame.
Synonymous substitution Hence, these lesions are called frameshift mutations.
These mutations cause the entire amino acid sequence
Thr Gln Ser Gly Base translationally downstream of the mutant site to bear
ACACAGAGTGGT substitutions no relation to the original amino acid sequence. Thus,
frameshift mutations typically result in complete loss of
Missense substitution normal protein structure and function.

Thr Ala Ala Trp Frameshift Now let’s turn to mutations that occur in regulatory
ACAGCAGCGTGGT mutation and other noncoding sequences (see Figure 14-3). Those
parts of a gene that do not directly encode a protein con-
Insertion tain many crucial DNA binding sites for proteins inter-
Regulatory site spersed among sequences that are nonessential to gene
expression or gene activity. At the DNA level, the dock-
ACACAGCGTGGT Point mutation ing sites include the sites to which RNA polymerase and
in noncoding its associated factors bind, as well as sites to which spe-
ACAGAGCGTGGT sequence cific transcription-regulating proteins bind. At the RNA
Regulatory protein level, additional important docking sites include the
cannot bind ribosome-binding sites of bacterial mRNAs, the 5Ј and 3Ј
splice sites for exon-joining in eukaryotic mRNAs, and
Figure 14-3 Consequences of point mutations within genes. In sites that regulate translation and localize the mRNA to
the top four panels, codons numbered 1 – 4 are located within particular areas and compartments within the cell.
the coding region of a gene.
It is much harder to predict the ramifications of mu-
Synonymous substitutions never alter the amino tations in parts of a gene other than the polypeptide-
acid sequence of the polypeptide chain. The severity of coding segments. In general, the functional consequences of
the effect of missense and nonsense mutations on the any point mutation in such a region depend on whether it
polypeptide differs from case to case. For example, if a disrupts (or creates) a binding site. Mutations that disrupt
missense mutation replaces one amino acid with a these sites have the potential to change the expression
chemically similar amino acid, referred to as a conserva- pattern of a gene by altering the amount of product ex-
tive substitution, then the alteration is less likely to af- pressed at a certain time or in a certain tissue or by alter-
fect the protein’s structure and function severely. Alter- ing the response to certain environmental cues. Such regu-
natively, chemically different amino acid substitutions, latory mutations will alter the amount of the protein
called nonconservative substitutions, are more likely to product produced but not the structure of the protein. Al-
produce severe changes in protein structure and func- ternatively, some binding-site mutations might completely
tion. Nonsense mutations will lead to the premature ter- obliterate a required step in normal gene expression (such
mination of translation. Thus, they have a considerable as the binding of RNA polymerase or splicing factors) and
effect on protein function. The closer a nonsense muta- hence totally inactivate the gene product or block its for-
tion is to the 3Јend of the open reading frame, the more mation. Figure 14-4 shows some examples of how differ-
plausible it is that the resulting protein might possess ent types of mutations affect mRNA and protein.
some biological activity. However, many nonsense muta-
tions produce completely inactive protein products. It is important to keep in mind the distinction
between the occurrence of a gene mutation — that is, a
Like nonsense mutations, indel mutations (base-pair change in the DNA sequence of a given gene — and
additions or deletions) have consequences on polypep- the detection of such an event at the phenotypic level.
Many point mutations within noncoding sequences elicit
little or no phenotypic change; these mutations fall
within sites that, for example, are between DNA bind-
ing sites for regulatory proteins. Such sites may be func-
tionally irrelevant, or other sites within the gene may
duplicate their function.

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14.1 Point mutations 457

Wild-type gene Missense mutation Nonsense mutation Frameshift mutation Regulatory region
(e.g., G и C→A и T) (e.g., CAA→TAA) (e.g., +A) mutation
mRNA
Protein NW NW NW No mRNA

NW No protein
NW

= Mutational site N = Northern blot W = Western blot = Unpredictable migration

(RNA) (protein)

Figure 14-4 Effects of common mutations. The effects of some common types of mutations
at the RNA and protein levels.

Mechanisms of point-mutation induction H H3C

When we examine the array of mutations induced by 5 N 5 O
different mutagens, we see that each mutagen is char- 64 64
acterized by a distinct mutational specificity, or “pref- H H H
erence,” both for a certain type of mutation (e.g., O
G · C : A · T transitions) and for certain mutational N1 2 3 N N1 2 3 N N
sites, called hot spots. Such mutational specificity was
first noted at the rII locus of the bacteriophage T4. O H N1 6 5 N H N
O N1 6 5
Mutagens act through at least three different mech- H 24 7 7
anisms. They can replace a base in the DNA, alter a base Cytosine N 3 Thymine 24
so that it specifically mispairs with another base, or dam- 8 3 8
age a base so that it can no longer pair with any base un- N 9 9
der normal conditions. N
N N
BASE REPLACEMENT Some chemical compounds are
sufficiently similar to the normal nitrogen bases of DNA H Guanine Adenine
that they are occasionally incorporated into DNA in
place of normal bases; such compounds are called base Figure 14-5 Pairing between the normal (keto) forms of the
analogs. Many of these analogs have pairing properties bases.
unlike those of the normal bases; thus they can produce
mutations by causing incorrect nucleotides to be in- rare. The imino or enol tautomer may pair with the
serted in the course of replication. To understand the ac- wrong base, forming a mispair. The ability of such a mis-
tion of base analogs, we must first consider the natural pair to cause a mutation in the course of DNA replica-
tendency of bases to assume different forms. tion was first noted by Watson and Crick when they for-
mulated their model for the structure of DNA (Chapter
Each of the bases in DNA can appear in one of sev- 7). Figure 14-6 demonstrates some possible mispairs re-
eral forms, called tautomers, which are isomers that dif- sulting from the change of one tautomer into another,
fer in the positions of their atoms and in the bonds be- termed a tautomeric shift.
tween the atoms. The forms are in equilibrium. The keto
form of each base is normally present in DNA (Figure Mispairs can arise spontaneously but can also
14-5), whereas the imino and enol forms of the bases are arise when bases become ionized. The mutagen
5-bromouracil (5-BU) is an analog of thymine that has
bromine at the carbon 5 position in place of the CH3
group found in thymine (Figure 14-7a). Its mutagenic
action is based on enolization and ionization. In 5-BU,
the bromine atom is not in a position in which it can
hydrogen-bond during base pairing. Thus the keto form
of 5-BU pairs with adenine, as would thymine; this pair-
ing is shown in Figure 14-7a. However, the presence of

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458 Chapter 14 • Mutation, Repair, and Recombination

H CH3

N O
H
HH O
N NN N
NN H
N
H N N
N O N
Thymine H N
O

Cytosine NN

H H3C Rare imino form H Rare enol form
of adenine (A*) of guanine (G*)
N O

HH H O
N
NN NN

H N H N
N N

O O

Rare imino form NN H N N
of cytosine (C*) Rare enol form N

Adenine of thymine (T*) H Guanine

Figure 14-6 Mismatched bases. Rare tautomeric forms of bases result in mismatches.

the bromine atom significantly alters the distribution of H
electrons in the base ring; so 5-BU can frequently change
to either the enol form or an ionized form; the enol and N O CH3 NH N
ionized forms of 5-BU pair with guanine (Figure 14-7b). N 87 N
5-BU causes G · C : A · T or A · T : G · C transi- 8 7 O
tions in the course of replication, depending on whether
5-BU has been enolized or ionized within the DNA N9 56 H N N9 56
molecule or as an incoming base. Hence the action of
5-BU as a mutagen is due to the fact that the molecule 4 1N 4 1N+ H
spends more of its time in the enol or ion form. N3 2
N3 2 N
Another widely employed mutagen is 2-amino-
purine (2-AP), an analog of adenine that can pair with NH O NH
thymine (Figure 14-8a). When protonated, 2-AP can
mispair with cytosine (Figure 14-8b). Therefore, when H H
2-AP is incorporated into DNA by pairing with
thymine, it can generate A · T : G · C transitions by 2-AP Thymine Protonated Cytosine
mispairing with cytosine in subsequent replications. Or, 2-AP
if 2-AP is incorporated by mispairing with cytosine, then (a)
G · C : A · T transitions will result when 2-AP pairs (b)
with thymine in subsequent replications. Genetic studies
have shown that 2-AP, like 5-BU, is highly specific for Figure 14-8 Alternative pairings for 2-aminopurine (2-AP).
transitions. This analog of adenine can pair with cytosine in its protonated
state (b).

H Br O– ON BASE ALTERATION Some mutagens are not incorpo-
rated into the DNA but instead alter a base, causing spe-
Br O H N N 56 NN cific mispairing. Certain alkylating agents commonly
N used as mutagens, such as ethylmethanesulfonate (EMS)
56 HN N 4 1N H and nitrosoguanidine (NG), operate by this pathway.
N N3 2 H N
4 1N H Such agents add alkyl groups (an ethyl group in the
N3 2 O Guanine case of EMS and a methyl group in the case of NG) to
many positions on all four bases. However, a mutation is
O most likely to occur when the alkyl group is added to
the oxygen at position 6 of guanine to create an O-6-
Common keto Adenine Rare alkylguanine. This alkylation leads to direct mispairing
form of 5-BU ionized form with thymine, as shown in Figure 14-9, and results in
G · C : A · T transitions in the next round of replica-
(a) of 5-BU tion. Alkylating agents can also modify the bases of in-
coming nucleotides in the course of DNA synthesis.
(b)
The intercalating agents are another important class
Figure 14-7 Alternative pairings for 5-bromouracil (5-BU). An of DNA modifiers. This group of compounds includes
analog of thymine, 5-BU can be mistakenly incorporated into proflavin, acridine orange, and a class of chemicals
DNA as a base. The ionized form base pairs with guanine. termed ICR compounds (Figure 14-10a). These agents
are flat planar molecules that mimic base pairs and are
able to slip themselves in (intercalate) between the
stacked nitrogen bases at the core of the DNA double

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14.1 Point mutations 459

H3C CH2 some repair mechanisms are themselves responsible for
mutating DNA. The name SOS comes from the idea that
NO O CH3 A•T this system is induced as an emergency response to pre-
NO vent cell death in the presence of significant DNA dam-
6 G•C age. As such, SOS induction is a mechanism of last
6 resort, a form of damage tolerance that allows the cell to
N 1NH 1 trade death for a certain level of mutagenesis.
N EMS N 1N H N3
NH N N It has taken over 30 years to figure how the SOS
H NH O system generates mutations while allowing DNA poly-
merase to bypass lesions at stalled replication forks. As
Guanine H Thymine you will see below, UV light usually causes damage to
O -6-Ethylguanine nucleotide bases in most organisms. An unusual class of
E. coli mutants that survived UV exposure without sus-
H3C CH3 taining additional mutations was isolated in the 1970s.
The fact that such mutants even existed suggested that
H3C O EMS H3C O O N some E. coli genes function to generate mutations when
T•A exposed to UV light.
1 4 4 C•G
N UV-induced mutation will not occur if the DinB,
N 3NH 1 3N H N N UmuC, or UmuDЈ genes are mutated. Recently it was
N discovered that these genes encode two error-prone
O DNA polymerases: DinB encodes DNA polymerase IV,
OHN while UmuC and UmuDЈ encode subunits of DNA
polymerase V. These polymerases overcome the block in
Thymine O -4-Ethylthymine H replication by adding nucleotides to the strand opposite
Guanine the damaged bases. Error-prone polymerases (also called
EP polymerases or sloppy copiers) have also been found in
Figure 14-9 Alkylation-induced specific mispairings. diverse taxa of eukaryotes from yeast to human, where
Treatment with EMS alters the structure of guanine and they contribute to a damage-tolerance mechanism called
thymine and leads to mispairings. translesion DNA synthesis that resembles the SOS bypass
system in E. coli. Figure 14-11 shows how these poly-
helix (Figure 14-10b). In this intercalated position, merases (called pol ␶ and pol ␩ in humans) function in
an agent can cause single-nucleotide-pair insertions or humans.
deletions.
Whereas error-prone polymerases always appear to
BASE DAMAGE A large number of mutagens damage be present in eukaryotic cells, they are induced by UV
one or more bases; so no specific base pairing is possible. exposure in E. coli. The first step in the SOS mechanism
The result is a replication block, because DNA poly- occurs when UV induces the synthesis of a protein called
merase cannot continue DNA synthesis past such a dam- RecA. We will see more of the RecA protein later in the
aged template base. In both prokaryotes and eukaryotes, chapter because it is a key player in many mechanisms of
such replication blocks can be bypassed by inserting non- DNA repair and recombination. When the replicative
specific bases. In E. coli, this process requires the activa-
tion of the SOS system. SOS and other mechanisms of
biological repair will be described later in this chapter.
However, an overview of this repair mechanism will be
presented in this section because, somewhat ironically,

H CH2CH2Cl
N

H3C CH3 CH2 Nitrogenous
CH2 bases

H CH2 Intercalated
N molecule

H2N N NH2 N N N OCH3 (b)

H3C CH3

Cl N

Proflavin Acridine orange ICR-191

(a)

Figure 14-10 Intercalating agents. (a) Structures of common intercalating agents and
(b) their interaction with DNA. [From L. S. Lerman, Proc. Natl. Acad. Sci. USA 39, 1963, 94.]

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460 Chapter 14 • Mutation, Repair, and Recombination

Stalled Base 1 Error-prone DNA Figure 14-11 Translesion DNA synthesis. Error-prone DNA
replication machine damage polymerases polymerases ␩ and ␶ allow the replication machine to get past
bind stalled a bulky damaged base, incorporating erroneous bases.
replication
machine. polymerase (DNA polymerase III) stalls at a site of DNA
damage, the DNA ahead of the polymerase continues to
pol η pol ι Location marker be unwound, exposing regions of single-stranded DNA
to show movement that become bound by single-strand-binding protein (SSB).
Next, RecA proteins join the SSB and form a protein –
2 Binding promotes DNA filament. The RecA filament is the biologically
active form of this protein. In this situation, RecA acts as
conformational a signal that leads to the induction of the error-prone
polymerase and attracts it to the stalled fork.
change.
Mutagens that create bases unable to form stable
Replication base pairs are thus dependent on SOS and similar sys-
tems for their mutagenic action, because the incorpora-
ηι begins. pol η tion of incorrect nucleotides requires the activation of
adds erroneous the SOS system. The category of SOS-dependent mu-
tagens is important because it includes most cancer-
bases. causing agents (carcinogens), such as ultraviolet (UV)
light and aflatoxin B1. Indeed, a great deal of work has
Erroneous pol η 3 Pol η been done on the relationship of mutagens to carcino-
bases disassociates. gens. The connections between mutation and cancer will
pol ι continues be discussed in detail in Chapter 17.
pol ι incorporating
erroneous bases
on strand
opposite
damaged base.

5′ 3′ 5′ O H O
OO H H2C N
OP O 2
O 3
O N1 6 4
OO
H2C OP 2 5
O
O N1 CH3
H2C
OO O 6 CH3
OP O
CH2 O 3 O
O O OO 54
H2C OP
PO N
O OO O
O 3′ H
OO CH2
OP N C O
3 2
O O C 4 5 6 1 N PO
H2C OO
C C
O H3C CH2
O
OO O
OP PO
O OO
O
H2C N C
3 2
O O C 4 5 6 1 N

C C
H3C
H2C
Figure 14-12 UV light – O O O H
generated photoproducts. N N
Photoproducts that unite Cyclobutyl ring
adjacent pyrimidines in O
DNA are strongly correlated
with mutagenesis. OO H CH3
OP H3C OH
[Left panel adapted from
E. C. Friedberg, DNA O O N N
Repair. Copyright 1985 H O O
by W. H. Freeman and 3′ H2C
Company. Right panel
from J. S. Taylor et al.] (a) Cyclobutane pyrimidine dimer O T (6-4) T
CH2
OO (b) 6-4 Photoproduct
O OP
PO
O
OO

5′

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14.2 Spontaneous mutation 461

Ultraviolet light generates a number of distinct types Guanine Aflatoxin B1
of alterations in DNA, called photoproducts, from the O
word photo for “light.” The most likely to lead to muta- O
tions are two different lesions that unite adjacent pyrim-
idines in the same strand. These lesions are the cyclobu- O
tane pyrimidine photodimer and the 6-4 photoproduct.
For the cyclobutane pyrimidine dimer, ultraviolet light HO
stimulates the formation of a four-membered cyclobutyl O
ring (shown in green in Figure 14-12a) between two ad-
jacent pyrimidines on the same DNA strand by acting HN N OO OCH3
on 5,6 double bonds. The 6-4 photoproduct structure
(Figure 14-12b) forms between the C-6 and C-4 posi- +
tions of two adjacent pyrimidines, most prevalently
5Ј-CC-3Јand 5Ј-TC-3Ј. The UV photoproducts signifi- H2N NN
cantly perturb the local structure of the double helix.
These lesions interfere with normal base pairing; hence, DNA
induction of the SOS system is required for mutagenesis. backbone
The incorrect bases are inserted across from UV photo-
products at the 3Јposition of the dimer. The C : T tran- Figure 14-13 The binding of metabolically activated aflatoxin
sition is the most frequent mutation, but UV light also B1 to DNA.
induces other base substitutions (transversions) and
frameshifts, as well as larger duplications and deletions. 14.2 Spontaneous mutation

Aflatoxin B1 (AFB1) is a powerful carcinogen origi- The origin of spontaneous hereditary change has always
nally isolated from peanuts infected with a fungus. Afla- been a topic of considerable interest. One of the first
toxin attaches to guanine at the N-7 position (Figure questions asked by geneticists was whether spontaneous
14-13). The formation of this addition product leads to mutations are induced in response to external stimuli, or
the breakage of the bond between the base and the whether variants are present at a low frequency in most
sugar, thereby liberating the base and resulting in an populations. An ideal experimental system to address
apurinic site (Figure 14-14). Studies of apurinic sites this important question was the analysis of mutations in
generated in vitro have demonstrated that the SOS by- bacteria that confer resistance to specific environmental
pass of these sites often leads to the insertion of an ade- agents not normally tolerated by wild types.
nine residue across from an apurinic site. Thus, agents
that cause depurination at guanine residues should tend Luria and Delbrück fluctuation test
to induce G · C : T · A transversions.
One experiment by Salvador Luria and Max Delbrück
MESSAGE Mutagens induce mutations by a variety of in 1943 was particularly influential in shaping our un-
mechanisms. Some mutagens mimic normal bases and are derstanding of the nature of mutation, not only in bacte-
ria, but in organisms generally. It was known at the time
incorporated into DNA, where they can mispair. Others damage that if E. coli bacteria are spread on a plate of nutrient
medium in the presence of phage T1, the phage soon in-
bases, which then are not correctly recognized by DNA fect and kill the bacteria. However, rarely but regularly,
colonies were seen that were resistant to phage attack;
polymerase during replication, resulting in mispairing. these colonies were stable and so appeared to be gen-
uine mutants. However, it was not known whether these

Aflatoxin

DNA DNA remains intact 0
A
Depurination 0 Replication
C G
G C
C

Figure 14-14 The loss of a purine residue from
a single strand of DNA.

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462 Chapter 14 • Mutation, Repair, and Recombination

mutants were produced spontaneously but randomly in 26. If the phage were inducing mutations, there was no
time or whether the presence of the phage induced a reason why fluctuation should be higher on the individ-
physiological change that caused resistance. ual cultures, because all were exposed to phage similarly.
The best explanation was that mutation was occurring
Luria reasoned that if mutations occurred sponta- randomly in time: the early mutations gave the higher
neously, then the mutations might be expected to occur numbers of resistant cells because they had time to pro-
at different times in different cultures; so the resulting duce many resistant descendants. The later mutations
numbers of resistant colonies per culture should show produced fewer resistant cells (Figure 14-15b).
high variation (or “fluctuation” in his word). He later
claimed that he obtained the idea while watching the This elegant analysis suggests that the resistant cells
fluctuating returns obtained by colleagues gambling on a are selected by the environmental agent (here, phage)
slot machine at a faculty dance in a local country club; rather than produced by it. Can the existence of mutants
hence the term “jackpot” mutation. in a population before selection be demonstrated di-
rectly? This demonstration was made possible by the use
Luria and Delbrück designed their “fluctuation test” of a technique called replica plating, developed by Joshua
as follows: They inoculated 20 small cultures, each with and Esther Lederberg in 1952. A population of bacteria
a few cells, and incubated them until there were 108 was plated on nonselective medium — that is, medium
cells per milliliter. At the same time, a much larger cul- containing no phage — and from each cell a colony grew.
ture also was inoculated and incubated until there were This plate was called the master plate. A sterile piece of
108 cells per milliliter. The 20 individual cultures and 20 velvet was pressed down lightly on the surface of the
samples of the same size from the large culture were master plate, and the velvet picked up cells wherever
plated in the presence of phage. The 20 individual cul- there was a colony (Figure 14-16). In this way, the velvet
tures showed high variation in the number of resistant picked up a colony “imprint” from the whole plate. On
colonies: 11 plates had 0 resistant colonies, and the re- touching the velvet to replica plates containing selective
mainder had 1, 1, 3, 5, 5, 6, 35, 64, and 107 per plate. medium (that is, containing T1 phage), cells clinging to
The 20 samples from the large culture showed much the velvet are inoculated onto the replica plates in the
less variation from plate to plate, all in the range of 14 to

(a) Physiological change Culture 2 Culture 3 Culture 4
Culture 1

(b) Random mutation Culture 2 Culture 3 Culture 4
Culture 1

Figure 14-15 “Fluctuation test” hypotheses. These cell pedigrees illustrate the expectations
from two contrasting hypotheses about the origin of resistant cells. [From G. S. Stent and

R. Calendar, Molecular Genetics, 2d ed. W. H. Freeman and Company, 1978.]

44200_14_p451-480 3/18/04 9:21 AM Page 463

14.2 Spontaneous mutation 463

Velvet surface after exposure to the selective agents, the patterns for
(sterilized) each plate would have been as random as the mutations
themselves. The mutation events must have occurred
Handle before exposure to the selective agent.

MESSAGE Mutation is a random process. Any allele in
any cell may mutate at any time.

Pressed on master plate Then pressed on replica Replica plating has become an important technique
containing grown colonies plate that distinguishes wild of microbial genetics. It is useful in screening for mu-
and mutant genotypes tants that fail to grow under the selective regime. The
position of an absent colony on the replica plate is used
Figure 14-16 Replica plating. Replica plating reveals mutant to retrieve the mutant from the master. For example,
colonies on a master plate through their behavior on selective replica plating can be used to screen auxotrophic mu-
replica plates. [From G. S. Stent and R. Calendar, Molecular tants in precisely this way. In general, replica plating is a
way of retaining an original set of strains on a master
Genetics, 2d ed. W. H. Freeman and Company, 1978.] plate while simultaneously subjecting replicas to various
kinds of tests on different media or under different envi-
same relative positions as those of the colonies on the ronmental conditions.
original master plate. As expected, rare resistant mutant
colonies were found on the replica plates, but the multi- Mechanisms of spontaneous mutations
ple replica plates showed identical patterns of resistant
colonies (Figure 14-17). If the mutations had occurred We now know that spontaneous mutations arise from a
variety of sources, including errors in DNA replication,
Master plate containing spontaneous lesions, and as we saw in Chapter 13, the
107 colonies of Ton s E. coli (T1-sensitive) insertion of transposable elements. Spontaneous muta-
tions are very rare, making it difficult to determine the
Replica plating underlying mechanisms. What sources of insight do we
then have into the processes governing spontaneous mu-
Plate 1 Plate 2 Plate 3 tation? Even though these mutations are rare, some
selective systems allow spontaneous mutations to be ob-
Series of replica plates containing high concentrations tained and then characterized at the molecular level —
of T1 phage and four Ton r colonies for example, their DNA sequences can be determined.
From the nature of the sequence changes, inferences can
Figure 14-17 Replica plating demonstrates the presence of be made about the processes that have led to the spon-
mutants before selection. The identical patterns on the replicas taneous mutations.
show that the resistant colonies are from the master. [From
SPONTANEOUS LESIONS Naturally occurring damage
G. S. Stent and R. Calendar, Molecular Genetics, 2d ed. to DNA, called spontaneous lesions, can generate muta-
W. H. Freeman and Company, 1978.] tions. Two of the most frequent spontaneous lesions are
depurination and deamination, the former being more
common.

We learned earlier that aflatoxin induces depurina-
tion, the loss of a purine base; however, depurination
also occurs spontaneously. A mammalian cell sponta-
neously loses about 10,000 purines from its DNA in a
20-hour cell-generation period at 37˚C. If these lesions
were to persist, they would result in significant genetic
damage because, during replication, the apurinic sites
cannot specify any kind of base, let alone the correct
one. However, as mentioned earlier in the chapter, under
certain conditions, a base can be inserted across from an
apurinic site, frequently resulting in a mutation.

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464 Chapter 14 • Mutation, Repair, and Recombination

HH O of mutations at each of a number of sites. The positions
N H of 5-methylcytosine residues correlate nicely with the
most mutable sites.
N Deamination N
Oxidatively damaged bases constitute a third type
(a) of spontaneous lesion that can lead to mutation. Active
oxygen species, such as superoxide radicals (O2·Ϫ ),
NO NO hydrogen peroxide (H2O2), and hydroxyl radicals
(·OH), are produced as by-products of normal aerobic
Cytosine Uracil metabolism. These oxygen species can cause oxidative
damage to DNA, as well as to precursors of DNA (such
HH as GTP), resulting in mutation. Such mutations have
been implicated in a number of human diseases. Figure
NO 14-20 shows two products of oxidative damage, one a
damaged thymidine residue, the other a damaged guano-
H3C 5 4 3 N H3C H sine residue. The 8-oxo-7-hydrodeoxyguanosine (8-oxo
(b) 6 2 Deamination N dG, or “GO”) product frequently mispairs with A, re-
sulting in a high level of G : T transversions.
1 O NO

N

5-Methylcytosine Thymine

Figure 14-18 Deamination of (a) cytosine and ERRORS IN DNA REPLICATION Mistakes made by
(b) 5-methylcytosine. the DNA replication apparatus are another source of
mutations.
The deamination of cytosine yields uracil (Figure 14-
18a). Unless corrected, uracil residues will pair with ade- Base substitutions No chemical reaction is perfectly
nine in the course of replication, resulting in the conver- efficient. Accordingly, an error in DNA replication can
sion of a G · C pair into an A · T pair (a G · C : A · T
transition). Deamination of 5-methylcytosine also occurs O O H
(Figure 14-18b). (Certain bases in prokaryotes and eu- CH3 N
karyotes are normally methylated.) The deamination of HN1 6 5
5-methylcytosine generates thymine (5-methyluracil). HN3 4 5 OH 7
Thus, C to T transitions generated by deamination are also
seen frequently at 5-methylcytosine sites. DNA sequence 26 OH 24 8O
analysis of hot spots for G · C : A · T transitions in 1
the lacI gene has shown that 5-methylcytosine residues O NH NH2 3 9
are present at the position of each hot spot. Some of the
data from this lacI study are shown in Figure 14-19. The N N
height of each bar on the graph represents the frequency
dR dR

Thymidine glycol 8-Oxo-7-hydrodeoxyguanosine
(8-oxo dG)

Figure 14-20 DNA damage products formed after attack by
oxygen radicals.

Number of occurrences 15 *

* GC AT

10 *

5*

0 50 100 150 200 250 300 350

Position

Figure 14-19 5-Methylcytosine hot spots in E. coli. Nonsense mutations at 15
different sites in lacI were scored. All resulted in G и C : A и T transition. The
asterisk (*) marks the positions of 5-methylcytosines, and the white bars mark sites
where transitions known to occur were not isolated in this group.[From C. Coulondre,

J. H. Miller, P. J. Farabaugh, and W. Gilbert, Nature 274, 1978,775.]

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14.2 Spontaneous mutation 465

www. ANIMATED ART Molecular mechanism of mutation Addition Deletion
Direction of DNA synthesis Direction of DNA synthesis

5' 9 CGTTTT 5' 9 CTGAGAGA
3' 9 GCAAAAACGTAC 9 3' 9 GACTCTCTCTCTGCA 9

Newly synthesized strand slips Template strand slips
Extra base loops out Extra bases loop out

T 5' 9 CT GAGAGA
5' 9 CG TTTT
3' 9 GC AAAAACGTAC 9 3' 9 GA CTCTCTCTGCA 9
CT
Loop stabilized by
repetitive sequences Loop stabilized by
repetitive sequences

T 5' 9 CT GAGAGAGACGT
5' 9 CG TTTTTGCATG
3' 9 GC AAAAACGTAC 9 3' 9 GA CTCTCTCTGCA 9
CT
Next round of
replication Next round of
replication

5' 9 CGTTTTTTGCATG 9 5' 9 CGTTTTTGCATG 9 5' 9 CTGAGAGAGACGT 9 5' 9 CTGAGAGAGAGACGT 9
3' 9 GCAAAAAACGTAC 9 3' 9 GCAAAAACGTAC 9 3' 9 GACTCTCTCTGCA 9 3' 9 GACTCTCTCTCTGCA 9
G • C and A • T base pairs deleted
A • T base pair added

Figure 14-21 A model for indel mutations resulting in frameshifts. dr ϭ deoxyribose.

occur when an illegitimate nucleotide pair (say, A и C) 9 GT CTGG CTGG CTGG CTGG C
forms in DNA synthesis, leading to a base substitution.
Also recall that earlier in the chapter we discussed that FS5, FS25, FS45, FS65
there are several forms of each base, called tautomers,
and that the change of one tautomeric form to another Wild type 5Ј 9 GT CTGG CTGG CTGG C 9 3Ј
can result in mispairing during DNA replication.
FS2, FS84
Base insertion and deletion Although some errors in
replication produce base-substitution mutations, other GT CTGG CTGG C
kinds of replication errors can lead to indel
mutations — that is, insertions or deletions of one or The majority of mutations at this site (represented here
more base pairs. When such mutations add or subtract by the mutations FS5, FS25, FS45, and FS65) result
a number of bases not divisible by three, they produce from the addition of one extra set of the four bases
frameshift mutations in the protein-coding regions. CTGG. A minority (represented here by the mutations
The nucleotide sequence at frameshift mutation hot FS2 and FS84) result from the loss of one set of the four
spots was determined in the lysozyme-encoding gene bases CTGG.
of phage T4. These mutations often occur at repeated
bases. The prevailing model (Figure 14-21) proposes How can we explain these observations? The model
that indels arise when loops in single-stranded regions predicts that the frequency of a particular indel depends
are stabilized by the “slipped mispairing” of repeated on the number of base pairs that can form during the
sequences in the course of replication. This mechanism slipped mispairing of repeated sequences. The wild-type
is sometimes called replication slippage. In the E. coli sequence shown for the lacI gene can slip out one
lacI gene, certain hot spots result from repeated se- CTGG sequence and stabilize this structure by forming
quences, just as predicted by this model. Figure 14-22 nine base pairs (apply the model in Figure 14-21 to the
depicts the distribution of spontaneous mutations in sequence shown for lacI). Whether a deletion or an in-
the lacI gene. Note how one site dominates the distrib- sertion is generated depends on whether the slippage
ution. In lacI, the major indel hot spot is a four-base- is on the template or on the newly synthesized strand,
pair sequence (CTGG) repeated three times in tan- respectively.
dem in the wild type (for simplicity, only one strand of
the DNA is shown): Larger deletions (more than a few base pairs) consti-
tute a sizable fraction of observed spontaneous muta-
tions, as shown in Figure 14-22. Most, although not all,

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466 Chapter 14 • Mutation, Repair, and Recombination

Point Mutations FS5, FS25,
FS45, FS65

S58 A6 S28 LacI gene
S114 75 100 125 150 175 200 225 250 275 300 325 350

25 50 S23 S10 FS2, S56 S42
Amino acid number S136 FS84 S24
Deletions S120

S74 S32
S112 S65

S86

Figure 14-22 The distribution of 140 spontaneous mutations in lacI. Boxes indicate position
of point mutations, with red designating fast-reverting mutants. Deletions are represented
in gold. Circles designate larger deletion and insertion mutants. Allele numbers correspond
to mutants that have been sequenced. [From P. J. Farabaugh, U. Schmeissner, M. Hofer,

and J. H. Miller, Journal of Molecular Biology 126, 1978, 847.]

of the deletions are of repeated sequences. Figure 14-23 been observed in many organisms. Like deletions, they
shows 9 deletions analyzed at the DNA sequence level often occur at sequence repeats.
in the lacI gene of E. coli. The results of further studies
have shown that the longer repeats constitute hot spots It must be noted that, in addition to their origin by
for deletions. Duplications of DNA segments have also replication slippage, deletions and duplications could be
generated by offset homologous recombination between
S74, S112 75 bases copies of the repeats.
C A A T T C A G G G T G G T G A A T G T G A A A C C ------C G C G T G G T G A A C C A G G
MESSAGE Spontaneous mutations can be generated by
Site Sequence No. of Occurrences several different processes. Spontaneous lesions and
(no. of bp) repeat replication errors generate most of the base-substitution and
bases deleted indel mutations, respectively.
20 to 95
146 to 269 GTGGTGAA 75 2 S74, S112 Spontaneous mutations in humans —
331 to 351 GCGGCGAT 123 1 S23 trinucleotide repeat diseases
316 to 338 AAGCGGCG 2 S10, S136
694 to 707 GTCGA 20 2 S32, S65 DNA sequence analysis has revealed the gene mutations
694 to 719 CA 22 1 S24 contributing to numerous human hereditary diseases.
943 to 956 CA 13 1 S56 Many are of the expected base-substitution or single-
322 to 393 G 25 1 S42 base-pair indel type. However, some mutations are more
658 to 685 None 13 1 S120 complex. A number of these human disorders are due to
None 71 1 S86 duplications of short repeated sequences.
27
A common mechanism responsible for a number of
Figure 14-23 Analysis of deletions in lacI in regions genetic diseases is the expansion of a three-base-pair
containing repeated sequences. For S74 and S112, one of two repeat. For this reason, they are termed trinucleotide
repeated sequences and all of the intervening sequences are repeat diseases. An example is the human disease called
deleted, producing the same final sequence. [From P. J. fragile X syndrome. This disease is the most common
form of inherited mental retardation, occurring in close
Farabaugh, U. Schmeissner, M. Hofer, and J. H. Miller, Journal of

Molecular Biology 126, 1978, 847.]

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14.2 Spontaneous mutation 467

to 1 of 1500 males and 1 of 2500 females. It is mani- The proposed mechanism for the generation of these
fested cytologically by a fragile site in the X chromo- repeats is a slipped mispairing in the course of DNA syn-
some that results in breaks in vitro. Fragile X syndrome thesis, just as discussed previously for the expansion of
results from changes in the number of a (CGG)n repeat the repeat at the lacI hot spot. However, the extraordinar-
in a region of the FMR-1 gene that is transcribed but not ily high frequency of mutation at the trinucleotide re-
translated (Figure 14-24a). peats in fragile X syndrome suggests that in human cells,
after a threshold level of about 50 repeats, the replication
How does repeat number correlate with the disease machinery cannot faithfully replicate the correct se-
phenotype? Humans normally show considerable varia- quence, and large variations in repeat numbers result.
tion in the number of CGG repeats in the FMR-1 gene,
ranging from 6 to 54, with the most frequent allele con- Other diseases, such as Huntington disease (see
taining 29 repeats. Sometimes, unaffected parents and Chapter 2), also have been associated with the expan-
grandparents give rise to several offspring with fragile X sion of trinucleotide repeats in a gene. Several general
syndrome. The offspring with the symptoms of the dis- themes apply to these diseases. In Huntington disease,
ease have enormous repeat numbers, ranging from 200 to for example, the wild-type HD gene includes a repeated
1300 (see Figure 14-24b). The unaffected parents and sequence, often within the protein-coding region, and
grandparents have also been found to contain increased mutation correlates with a considerable expansion of
copy numbers of the repeat, but ranging from only 50 to this repeat region. The severity of the disease correlates
200. For this reason, these ancestors have been said to with the number of repeat copies.
carry premutations. The repeats in these premutation al-
leles are not sufficient to cause the disease phenotype, but Huntington disease and Kennedy disease (also called
they are much more unstable (i.e., readily expanded) than X-linked spinal and bulbar muscular atrophy) result from
normal alleles, and so they lead to even greater expansion the amplification of a three-base-pair repeat, CAG. Nor-
in their offspring. (In general, it appears that the more ex- mal persons have an average of 19 to 21 CAG repeats,
panded the repeat number, the greater the instability.) whereas affected patients have an average of about 46.
In Kennedy disease, which is characterized by progressive

(CGG)n TAA
(a) ATG

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
38 kb

(b) Phenotype Transmission Methylation Transcription
FMR-1 gene
Normal Stable No Yes
Normal

. . .(CGG)6–59 ATG

Premutation ATG Largely Unstable, No Yes
(CGG)60–200 normal prone to Yes No
... expansion

Full mutation (CGG)>200 ATG Affected Unstable

X ...

Figure 14-24 The FMR-1 gene involved in fragile X syndrome. (a) Exon structure and upstream
CGG repeat. (b) Transcription and methylation in normal, premutation, and full mutation
alleles. The red circles are methyl groups. [W. T. O’Donnell and S. T. Warren, Ann. Rev. Neuroscience

25, 2002, 315 – 338, Figure 1.]

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468 Chapter 14 • Mutation, Repair, and Recombination

muscle weakness and atrophy, the expansion of the • The repair pathway deletes the damaged DNA and
trinucleotide repeat occurs in the gene that encodes the uses an existing complementary sequence as a
androgen receptor. template to restore the normal sequence.

Properties common to some trinucleotide-repeat Direct reversal of damaged DNA
diseases suggest a common mechanism by which the ab-
normal phenotypes are produced. First, many of these The most straightforward way to repair a lesion is to
diseases seem to include neurodegeneration — that is, reverse it directly, thereby regenerating the normal base
cell death within the nervous system. Second, in such (Figure 14-25a). Although some types of damage are
diseases the trinucleotide repeats fall within the open essentially irreversible, in a few cases lesions can be re-
reading frames of the transcripts of these genes, leading paired by direct reversal. One case is a mutagenic photo-
to expansions or contractions of the number of repeats dimer caused by UV light. The cyclobutane pyrimidine
of a single amino acid in the polypeptide (for example, photodimer can be repaired by an enzyme called a pho-
CAG repeats encode a polyglutamine repeat). Thus, it is tolyase. The enzyme binds to the photodimer and splits
no accident that these diseases entail expansions of it, in the presence of certain wavelengths of visible light,
codon-size three-base-pair units. to regenerate the original bases (Figure 14-26). This re-
pair mechanism is called light repair or photorepair. The
But this explanation cannot hold for all trinucleotide- photolyase enzyme cannot operate in the dark, and so
repeat diseases. After all, in fragile X syndrome, the other repair pathways are required to remove UV dam-
trinucleotide expansion occurs near the 5Ј end of the age in the absence of visible light.
FMR-1 mRNA, before the translation start site. Thus, we
cannot ascribe the phenotypic abnormalities of the Alkyltransferases also are enzymes that directly re-
FMR-1 mutations to an effect on protein structure. One verse lesions. They remove certain alkyl groups that have
clue to the problem with the mutant FMR-1 genes is been added to the O-6 positions of guanine (see Figure
that they, unlike the normal gene, are hypermethylated, 14-9) by such mutagens as nitrosoguanidine and ethyl-
a feature associated with transcriptionally silenced genes methanesulfonate. The methyltransferase from E. coli has
(see Figure 14-24b). Based on these findings it is hy- been well studied. This enzyme transfers the methyl
pothesized that repeat expansion leads to changes in
chromatin structure that silence transcription of the mu- (a) Direct reversal
tant gene (see Chapter 10). In support of this model is
the finding that the FMR-1 gene is deleted in some pa- (b) Base excision and replacement
tients with fragile X syndrome. These observations sup-
port a loss-of-function mutation. (c) Segment removal and replacement

MESSAGE Trinucleotide-repeat diseases arise through the Figure 14-25 Three types of repair of DNA with a damaged
expansion of the number of copies of a three-base-pair base.
sequence normally present in several copies, often within
the coding region of a gene.

14.3 Biological repair mechanisms

Living cells have evolved a series of enzymatic systems
that repair DNA damage in a variety of ways. The low
rate of spontaneous mutation is indicative of the effi-
ciency of these repair systems. We can think of the spon-
taneous mutation rate as being at a balance point be-
tween the rate at which premutational damage arises
and the rate at which repair systems recognize this dam-
age and restore the normal base sequence. Failure of
these systems can lead to a higher mutation rate, as we
shall see later.

Let’s now examine some of the repair pathways, be-
ginning with error-free repair. For this pathway one of
two things can happen:

• The repair pathway chemically repairs the damage to
the DNA base.

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14.3 Biological repair mechanisms 469

UV-induced photodimers Thymine Base excision repair Base excision repair (see Figure
14-25b) is carried out by DNA glycosylases that cleave
www. ANIMATED ART and excision repair Thymine base – sugar bonds, thereby liberating the altered bases
and generating apurinic or apyrimidinic sites. An enzyme
Photolyase DNA backbone called AP endonuclease then cuts the sugar-phosphate
+ Photodimer backbone around the site lacking a base. A third enzyme,
deoxyribophosphodiesterase, cleans up the backbone
white UV light by removing a stretch of neighboring sugar-phosphate
light residues so that a DNA polymerase can fill the gap with
nucleotides complementary to the other strand. DNA
Figure 14-26 Induction and removal of UV-induced pyrimidine ligase then seals the new nucleotide into the backbone
photodimer. (Figure 14-27).

group from O-6-methylguanine to a cysteine residue on Numerous DNA glycosylases exist. One, uracil –
the protein. However, the transfer inactivates the en- DNA glycosylase, removes uracil from DNA. Uracil
zyme, so this repair system can be saturated if the level residues, which result from the spontaneous deamination
of alkylation is high enough. of cytosine (see Figure 14-18), can lead to a C : T tran-
sition if unrepaired. One advantage of having thymine
Homology-dependent repair systems (5-methyluracil) rather than uracil as the natural pairing
partner of adenine in DNA is that spontaneous cytosine
One of the overarching principles guiding cellular ge- deamination events can be recognized as abnormal and
netic systems is the power of nucleotide sequence com- then excised and repaired. If uracil were a normal con-
plementarity. (You will recall that genetic analysis also stituent of DNA, such repair would not be possible.
depends heavily on this principle.) Important repair
systems exploit the properties of antiparallel comple- AP endonuclease
mentarity to restore damaged DNA segments back to makes cut
their initial, undamaged state. In these systems, a seg-
ment of a DNA chain is removed and replaced with a Excision
newly synthesized nucleotide segment complementary exonuclease removes
to the opposite template strand. Because these systems stretch of DNA
depend on the complementarity, or homology, of the
template strand to the strand being repaired, they are Polymerase
called homology-dependent repair systems. Because re- synthesizes
pair takes place through a template, the rules of DNA new DNA
replication ensure that repair is accomplished with
high fidelity — that is, it is error-free. There are two ma- Ligase
jor homology-dependent error-free repair systems. One seals
system (excision repair) repairs damage that has been nick
detected before replication. The other (postreplication
repair) repairs damage that is detected in the course of Figure 14-27 Repair of AP (apurinic or apyrimidinic) sites. AP
the replication process or afterward. endonucleases recognize AP sites and cut the phosphodiester
bond. A stretch of DNA is removed by an exonuclease, and the
EXCISION-REPAIR PATHWAYS Unlike the examples of resulting gap is filled in by DNA polymerase I and DNA ligase,
reversal of damage described above, excision repair en- using the complementary strand as template. [After B. Lewin,
tails the removal and replacement of an entire base.
Genes. Copyright 1983 by John Wiley.]

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470 Chapter 14 • Mutation, Repair, and Recombination

5' 3' the chapter, several human diseases are caused by muta-
tions in some of the genes that encode repair proteins.
3' 5'
uvrABC excinuclease removes a In yeast, nucleotide excision repair is carried out by
12-nucleotide fragment of DNA the multisubunit repairosome, a complex made up of
more than 20 different polypeptides. The repairosome is
DNA polymerase I able to recognize damaged DNA, excise about 30 nu-
synthesizes new DNA cleotides around the damage, and fill in the gap using
the complementary strand as template (Figure 14-29). It
Ligase joins has been noted that this system preferentially repairs the
DNA segments template (transcribed) DNA strand. How can the re-
pairosome “know” which strand of a gene is transcribed
Figure 14-28 Nucleotide excision repair. Repair of a region and which is not? One clue is that seven of the polypep-
of DNA containing a thymidine dimer. The thymidine dimer is tides of the repairosome are also subunits of the basal
shown in blue, and the new region of DNA is shown in red. transcription apparatus that is able to discriminate be-
tween the template and nontemplate strands of DNA
Nucleotide excision repair Base excision repair can cor- (described in Chapter 8). One model is that the pres-
rect only damaged bases that can be removed by a specific ence of DNA damage leads to the dissociation of the
DNA glycosylase. However, there are more ways to dam- basal transcription apparatus and the assembly of the
age a base than there are glycosylases to remove them. repairosome. Presumably, in this way, a mutant gene will
Thus, another system is required to repair the damage that be repaired before it can be transcribed.
the glycosylases cannot. Rather than recognize a particular
damaged base, the nucleotide excision-repair system de- Repairosome
tects distortions in the double helix caused by the pres-
ence of an abnormal base (see Figure 14-25c). Mutations 5' 3'
that cause such distortions include the pyrimidine dimers 3' 5'
caused by UV light and the addition of aflatoxin to gua- TFIIH subunit
nine residues. Detection of a distortion initiates a multistep unwinds DNA Formation of a
repair process involving many proteins. In E. coli, a com- bubble
plex consisting of three enzymatic activities encoded by Rad3
the uvrABC genes detects the distortion and cuts the dam-
aged strand at two sites flanking the lesion (Figure 14-28). Ssl2
The uvrABC exinuclease, as it is called, excises precisely
12 nucleotides: 8 from one side of the damage and 4 from Excision of
the other side. The 12-nucleotide gap is then filled by damaged strand
DNA polymerase I, using the template strand to produce
an accurate copy of the original DNA sequence. DNA 3' incision 5' incision
ligase then seals the new oligonucleotide into place.
DNA synthesis
MESSAGE Base excision repair and nucleotide excision and ligation
repair are error-free repair mechanisms that recognize and
remove mispaired bases prior to replication and utilize the Figure 14-29 Nucleotide excision repair in eukaryotes. In this
undamaged DNA complementary strand to guide repair. example a pyrimidine dimer (triangle) causes a bulge, which is
recognized by a repairosome. Various proteins in the
Transcription-coupled repair in eukaryotes As in pro- repairosome make a single-stranded bubble, and one single
karyotes, eukaryotic error-free repair involves distinct strand is cut out and resynthesized. [Adapted from Encyclopedia
base excision- and nucleotide excision-repair systems.
DNA repair systems in eukaryotes are highly conserved of Life Sciences, 2001, E. C. Friedberg, “Nucleotide Excision Repair
from yeast to humans, and for this reason, yeast has
again proved to be a useful model. As we will see later in in Eukaryotes,” Figure 1.]

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14.3 Biological repair mechanisms 471

Why is it important to couple transcription and re- 5' Me 3'
pair? Unlike actively dividing E. coli and most other 3' G GATC 5'
prokaryotes, most of the cells in a multicellular organism C CTAG
are terminally differentiated and no longer dividing, so 5'
replication repair is not possible. These cells are not 3' Me
dead, however; their genes are being actively transcribed
into mRNAs, which are translated into proteins. DNA 5' G Me 3'
that is damaged in these cells must also be repaired, 3' T CGTAATGC 5'
since mutations in certain genes could have devastating Enzyme complex
consequences to the health of the whole organism. As recognizes mismatch G GCATTACG
you will see in Chapter 17, most cancerous tumors de- in hemimethylated C
velop from somatic cells that have sustained mutations DNA
that have not been repaired. 5' Me
3'
POSTREPLICATION REPAIR Some repair pathways are Excision of mismatched
capable of recognizing errors that usually occur during 5' base on unmethylated strand
replication but fail to be corrected by the 3Ј-to-5Јproof- 3' and resynthesis
reading function of the replicative polymerase. One such Repair synthesis
pathway, termed the mismatch-repair system, can detect and full methylation Me 3'
such mismatches. Mismatch-repair systems have to do at of DNA 5'
least three things: 5' G GATC
3' C CTAG
1. Recognize mismatched base pairs
Me
2. Determine which base in the mismatch is the
incorrect one Me 3'
G GATC 5'
3. Excise the incorrect base and carry out repair C CTAG
synthesis
Me
The second property is the crucial one of such a sys-
tem. Unless it is capable of discriminating between the Figure 14-30 Model for mismatch repair in E. coli. DNA is
correct and the incorrect bases, the mismatch-repair sys- methylated at A residues in the sequence GATC. DNA
tem cannot determine which base to excise to prevent a replication yields a hemimethylated duplex that exists until
mutation from arising. If, for example, a G и T mis- methylase can modify the newly synthesized strand. The
match occurs as a replication error, how can the system mismatch-repair system makes any necessary corrections
determine whether G or T is incorrect? Both are normal based on the sequence found on the methylated strand
bases in DNA. But replication errors produce mis- (original template). [After E. C. Friedberg, DNA Repair. Copyright
matches on the newly synthesized strand, and so the re-
pair system knows that it is the base on this strand that 1985 by W. H. Freeman and Company.]
must be recognized and excised.
ing, and it provides a convenient tag that can be de-
The mismatch-repair system is best characterized in tected by other enzyme systems. Figure 14-30 shows
bacteria. Recall from Chapter 10 that bacterial DNA is the replication fork during mismatch correction. Note
methylated; this methylation normally takes place after that only the old strand is methylated at GATC se-
replication. To distinguish the old, template strand from quences right after replication. After the mismatched
the newly synthesized strand, the bacterial repair system site has been identified, the mismatch-repair system
takes advantage of a delay in the methylation of the fol- corrects the error.
lowing sequence:
The mismatch-repair system has also been charac-
5Ј-G-A-T-C-3Ј terized in humans. Figure 14-31 depicts a model of
3Ј-C-T-A-G-5Ј how the human mismatch-repair system carries out the
correction. An important target of the human mismatch-
The methylating enzyme is adenine methylase, which repair system is short repeat sequences that can be
creates 6-methyladenine on each strand. However, it expanded or deleted during replication by the slipped-
takes adenine methylase several minutes to recognize mispairing mechanism described previously (see Figure
and modify the newly synthesized GATC stretches. 14-21). Mutations in some of the components of this
During that interval, the mismatch-repair system can pathway have been shown to be responsible for several
operate because it can now distinguish the old strand human diseases, especially cancers. There are thousands
from the new one by the methylation pattern. Methy- of short repeats (microsatellites) located all over the
lating position 6 of adenine does not affect base pair- genome. Although most are located in noncoding re-
gions (since most of the genome is noncoding), a few
are located in genes that are critical for normal growth
and development.

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472 Chapter 14 • Mutation, Repair, and Recombination

TCA GGGG GTGTGT as a template for the DNA synthesis needed to fill the
AGT CCCC CACACA single-strand gap. However, what would happen if both
strands of the double helix were damaged in such a way
Replication error that complementarity could not be exploited? One way
this might happen is if both strands of the double helix
TCG CACACA Extrahelical GT were to break at sites that were close together. A mutation
G loop like this is called a double-strand break. If left unrepaired,
AGT G GGG GT GT double-strand breaks can cause a variety of chromosomal
Mispair C CCC CACACA aberrations resulting in cell death or a precancerous state.

Slippage Interestingly, the ability of double-strand breaks to ini-
tiate chromosomal instability is an integral feature of some
hMutSα recognizes and normal cellular processes that require DNA rearrange-
binds to site of error. ments. One example is the generation of the diversity of
antibodies in the cells of the mammalian immune system.
hMutSα Another is meiotic recombination, which uses double-
strand breaks to generate genetic diversity. As will be seen
ATP hMutLα is recruited. in the remainder of this chapter, the cell uses many of the
ADP same proteins and pathways to repair double-strand breaks
hMutLα and to carryout meiotic recombination. For this reason,
P we begin by focusing on the molecular mechanisms that
repair double-strand breaks before turning our attention
CNA to the mechanism of meiotic recombination.

Exonuclease removes Double-strand breaks can arise spontaneously (for
the mismatched segment. example, in response to reactive oxygen species), or
they can be induced by ionizing radiation. Two distinct
AGT CCCC CACACA mechanisms are used to repair these potentially lethal
lesions: nonhomologous end joining and homologous
DNA polymerase synthesizes recombination.
the excised region.
NONHOMOLOGOUS END-JOINING As mentioned earlier,
DNA ligase joins the newly DNA repair is important to prevent precancerous muta-
synthesized DNA to the tions from occurring in the nondividing cells of multicel-
original strand. lular organisms. However, when a double-strand break
occurs in cells that have stopped dividing, error-free repair
TCA GGGG GTGTGT is not possible because neither of the two usual sources
AGT CCCC CACACA of undamaged DNA is available as a template for new
DNA synthesis. That is, complementarity cannot be ex-
Figure 14-31 Model for mismatch repair in humans. Errors ploited because both strands of the DNA helix are dam-
arising at replication, such as mispaired regions and loops aged and, in the absence of replication, there is no sister
from replication slippage, can be removed and repolymerized chromatid. However, as was the case for the error-prone
by the proteins shown. [Adapted from Encyclopedia of Life translesion synthesis (including the SOS system in E. coli),
the consequences of imperfect repair may be less harmful
Sciences, 2001, P. Karran, “Human Mismatch Repair: Defects and to the cell than leaving the lesion unrepaired. In this case,
it is better to put the free ends back together so they can-
Predisposition to Cancer,” Figure 1.] not initiate chromosomal rearrangements, even if this
means that some sequence may be lost. Putting the ends
MESSAGE The mismatch-repair system corrects errors back together is accomplished by a mechanism called
in replication that are not corrected by the proofreading nonhomologous end-joining, which involves the three steps
function of the replicative DNA polymerase. Repair is shown in Figure 14-32. These steps include the binding of
restricted to the newly synthesized strand, which is recognized the broken ends by 3 proteins (KU70, KU80, and a large
by the repair machinery in prokaryotes because it lacks a DNA-dependent protein kinase) followed by the trim-
methylation marker. ming of the ends so that they can be ligated together. In
mammals, several of the proteins in this pathway also
Repair of double-strand breaks participate in the end-joining reactions associated with
the programmed rearrangements of antibody genes.
As we have seen, DNA complementarity is an important
resource that is exploited by many error-free correction
systems. Such error-free repair is characterized by two
stages: (1) removal of damaged and nearby DNA from one
strand of the double helix and (2) use of the other strand

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14.4 The mechanism of meiotic crossing-over 473

Double-strand break Double-strand break

Protein kinase Sister
KU80 and KU70 chromatids

End-binding by Trimming of 5' ends
protein complex
RAD51
3'
5' 3'

Formation of DNA–protein filament

Trimming of ends

Homology search
Strand invasion forms a short DNA joint

Joining of ends (ligation)

DNA polymerase DNA synthesis

Figure 14-32 Mechanism of nonhomologous end-joining of
double-strand breaks. This is an error-prone mechanism.

HOMOLOGOUS RECOMBINATION The mechanism of DNA ligase Sealing of strands
homologous recombination utilizes the sister chromatid
to repair double-strand breaks. For this reason, repair is Figure 14-33 Repair of a double-strand break by homologous
usually error-free. The mechanism of homologous re- recombination. A double-strand break induces an enzyme to
combination is shown in Figure 14-33. Key steps are the chew back 5Ј ends, leaving 3Ј overhangs that are coated with
binding of the broken ends by specialized proteins and proteins, including RAD51, a RecA homolog. A segment of the
enzymes, the trimming of the 5Ј ends to expose single- sister chromatid (blue) is used as a template to repair the
stranded regions, and the coating of these regions with break. [From D. C. van Gent, J. H. J. Hoeijmakers, and R. Kandar,
proteins that include the RecA homolog, RAD51. Recall
that during the SOS response, RecA monomers associate Nature Reviews: Genetics 2, 2001, 196 – 206.]
with regions of single-stranded DNA to form long heli-
cal filaments. Similarly, RAD51 forms long filaments as 14.4 The mechanism
it associates with the exposed single-stranded region. of meiotic crossing-over
The RAD51 – DNA filament then takes part in a re-
markable search of the undamaged sister chromatid for Our discussion of double-strand breaks leads naturally
the complementary sequence that will be used as a tem- into the topic of crossing-over at meiosis. This is because,
plate for DNA synthesis. Once the complementary re- according to the current molecular model of crossing-
gion is found, a joint molecule forms between homolo- over, a double-strand break initiates the crossover event.
gous damaged and undamaged duplex DNAs. The The molecular details of the crossover process will seem
sequences missing from the damaged strand are then very familiar from the above discussion of double-strand
copied from the complementary sister chromatid. break repair.

MESSAGE Double-strand breaks are extremely dangerous Crossing-over is a remarkably precise process. It
because they can lead to chromosome rearrangements that takes place between two homologous nonsister chro-
result in cell death or aberrant growth and development. matids. Some kind of cellular machinery takes these two
Nondividing cells efficiently join the ends of double-strand huge molecular assemblages, breaks them at the same
breaks with an error-prone process called nonhomologous end- relative position, and then rejoins them in a new arrange-
joining. Dividing cells utilize homologous recombination ment so that no genetic material is lost or gained in either.
whereby the free ends invade the homologous region of the
sister chromatid to initiate DNA synthesis and error-free repair.

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474 Chapter 14 • Mutation, Repair, and Recombination

The molecular mechanism is thought to comprise two produce identical sister genotypes.) Nonidentical sister
key steps: spore genotypes must have arisen from heteroduplex
DNA in the meiotic product, that is, DNA with a seg-
1. A double-strand break. One key piece of evidence ment in which one strand is the nucleotide sequence of
here was that in yeast transformation the the A allele and one strand is the nucleotide sequence of
incorporation of a circular plasmid into the yeast the a allele. After mitosis the two sister cells resulting
genome is stimulated 1000-fold when the plasmid is from division of such a heteroduplex-containing nucleus
cut to become linear. Broken DNA ends seem to be will be different, one with A and one a.
recombinogenic; that is, they promote recombination.
A
2. The formation of heteroduplex DNA. This is a A
hybrid type of DNA molecule that is composed of a A
single DNA strand from a chromatid derived from a
one parent, and a single strand from a chromatid
derived from the other parent. a
a
The first evidence for heteroduplex DNA was also
provided by genetics, specifically ascus analysis. Octads Let’s assume that A and a differ by a single base
are particularly informative in pointing to the existence pair; that base pair is G и C in A and A и T in a. Let’s
of heteroduplexes in crossing-over. We saw in Chapter further assume that for any particular gene, rarely a sin-
2 that in fungi a cross A ϫ a will create a monohybrid gle heteroduplex is formed at meiosis such that the
meiocyte Aրa that is expected to segregate in a 1 : 1 ra- A и T in one of the products becomes G и T (we will
tio in the meiotic products according to the law of see the mechanism soon). Then we can represent the
equal segregation. Indeed the 1 : 1 ratio is found in products of meiosis as
most fungal meiocytes: we see 4 A and 4 a. However, in
rare meiocytes (generally on the order of 0.1 to 1 per- 1. G и C 2. G и C 3. G и T 4. A и T
cent) any one of four types of aberrant ratios can be
found, and these give the clues needed to build a hetero- After the postmeiotic mitosis, the resulting octad
duplex crossover model. The aberrant ratios are as will be
follows:
1. G и C 2. G и C 3. G и C 4. G и C
Sac 5. G и C 6. A и T 7. A и T 8. A и T

AA a AA which is the observed 5 : 3 ratio.
Spore pair 1 However, from what we have learned in this chap-

AA a AA ter, we know that the G и T of the heteroduplex is a
candidate for mismatch repair. Let’s assume that such a
AA a AA repair system excises the T of the GT heteroduplex and
Spore pair 2 inserts C, so we get GC. In this ascus there will be a 3 : 1
ratio of GC : AT and the octad will show a 6 : 2 ratio.
AA aAa
aAaAa In meioses that produce aberrant ratios, it was ob-
Spore pair 3 served that there is a crossover between flanking genes
aAa a a at much higher frequencies than expected. Hence it
a aAaa seemed likely that heteroduplex formation might be
Spore pair 4 part of the normal crossover process. Furthermore, pre-
a aAaa sumably by chance the heterozygous gene under study
rarely happened to be in the middle of the molecular
4:4 6:2 2:6 5:3 3:5 events of a crossover, and it was these molecular events
that led to the gene conversion. Putting all these ideas
Norm Aberrant ratios together led to the double-strand break model, one of
several heteroduplex models of crossing-over.
These asci all have more than 4 copies of one geno-
type; this result is unexpected based on Mendel’s first The model is shown in Figure 14-34. In one a chro-
law of equal segregation. The one or two “extra” cases matid a double-strand break occurs, and erosion of the
are said to have undergone gene conversion from wild ends results in short regions of single-stranded DNA.
type to mutant or mutant to wild type. All the aberrant The 3Ј end of one of these strands “invades” an A chro-
ratios need to be explained, but we will concentrate on matid. The invader primes synthesis of its missing bases,
two, the 5 : 3 and the 6 : 2, because their explanation using the antiparallel strand of the A chromatid as a
embodies the same elements as all. template. This new synthesis displaces a single-stranded

The 5 : 3 ratio is particularly interesting because in
this octad there is a pair of nonidentical sister spores. (Re-
call that the postmeiotic round of mitosis is expected to

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14.4 The mechanism of meiotic crossing-over 475

a inner chromatid a
DNA double helix
A inner chromatid
a
1 Double-strand break A
and erosion of ends
DNA double helix
A

a 3' ends

A

A Figure 14-34 The
a double-strand break model
of crossing-over. Only the
2 Broken end invasion A two “inner” chromatids
and loop formation are shown of a hetero-
A zygous meiocyte A/a.
3 Migration of loop a A product of the model is
and synthesis a heteroduplex region. In
A this illustration, by chance
4 Gap-filling and A the heteroduplex spans
union of free ends the A/a mutant site.
1 A Holliday Mismatch repair causes
22 a 3 structure gene conversion, to A in
the case shown.

A 4 4
A

1 A
3
Heteroduplex 5a Breakage and reunion 5b Breakage and reunion
at 2 and 4
with mismatch at 2 and 3
a
a

AA
AA

AA

6 Repair to A 6 Repair to A
A A

AA
AA

A A

Full double-helical Noncrossover
crossover outcome
This will be
This will be
a 6:2 octad. a 6:2 octad.

If no A/a repair,
a 5:3 octad would result.

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476 Chapter 14 • Mutation, Repair, and Recombination

loop, which hybridizes to the noninvading a single edly evolved for mutation repair, they have been put to
strand, thus forming a small “Aa” heteroduplex region, new uses.
and serves as a template to restore the missing bases on
that strand. Filling in of gaps by polymerase activity and MESSAGE A double-strand break on one chromatid leads
joining of DNA ends by ligase result in a peculiar struc- to two single-stranded Holliday junctions surrounding a
ture that looks like two single-stranded crossovers. Note heteroduplex region. This overall molecular structure can be
that this structure also contains the single heteroduplex resolved into a single reciprocal double-stranded crossover.
that we need. Repair or nonrepair of the heteroduplex results in a 6 : 2 or a
5 : 3 ratio.
The single-stranded “crossovers” are called Holliday
structures, after Robin Holliday, who first proposed them Recombination between alleles of a gene
in the 1960s. Physical (as opposed to genetic) evidence
for Holliday structures has been obtained independently. In a cross between two mutant alleles, say, aЈ ϫ aЈЈ, the
They are unstable and must be resolved in one of two heteroduplex might span one mutant site, or both mu-
ways. Simply put, they can each be resolved by either tant sites. Such a heteroduplex would be as follows:
“vertical” or “horizontal” breakage and reunion of single
strands, as shown in the figure. One resolution results in a ----aϩ---- ϩ ----
double-stranded reciprocal crossover (shown at left) and ---- ϩ ----aϪ----
the other a noncrossover (right). Hence the association
with crossovers is explained. Note that if the hetero- Mismatch repair of at least one site to the wild-type
duplex formed via a double-strand break on the other sequence would result in a wholly wild-type strand.
chromatid, we could explain the 3 : 5 and 2 : 6 ratios, too. These intragenic recombinants can be detected in sev-
eral ways. In crosses between auxotrophic microorgan-
Overall, note again the use of molecular mechanisms isms, they would be prototrophs, and thus they can be
in crossing-over that seem “stolen” from the mutation detected most simply by plating on minimal medium.
and repair processes. Although these enzymes undoubt-

KEY QUESTIONS REVISITED • Can a cell repair mutations?

• What is the molecular nature of mutations? Yes. Some repair systems are very efficient at restoring
the original sequence. Others are error-prone: they
A large proportion of mutations are changes to, addi- convert the original chemical event into a permanent
tions to, or losses of one or a small number of bases in mutation.
the gene sequence. Such changes may cause a change of
amino acid (missense mutations) or produce a new stop • What is the molecular mechanism of crossing-over?
codon (nonsense mutations). Additions or losses can
cause frameshift mutations. In the current best model, a double-stranded break in
one chromatid initiates formation of a pair of single-
• How do certain types of radiation and chemicals stranded Holliday junctions that can be resolved to
cause mutation? form one standard double-helical crossover. Hetero-
duplex DNA is formed in the process, and if it hap-
By acting on and chemically changing the DNA. Com- pens to span a heterozygous gene, aberrant ascus ratios
mon changes are base replacement, base alteration, or can result.
base damage.
• Do mutational repair systems participate in crossing-
• Are induced mutations different from spontaneous over?
mutations?
Yes, certain steps in the double-strand model (such as
Spontaneous mutations are largely the product of errors double-strand breaks, exonuclease activity, mismatch re-
made by cellular enzymes. A wide spectrum of changes pair, polymerase activity, ligase activity) are very similar
can result. Some mutagens can also produce a wide to several types of mutational repair. Common enzymes
spectrum of changes, but many produce a preponder- are also involved.
ance of a certain type such as G · C-to-A · T transition
substitutions.

SUMMARY scription termination) codons. A purine replaced by the
other purine (or a pyrimidine replaced by the other
DNA change within a gene (point mutation) generally pyrimidine) is called a transition. A purine replaced by a
involves one or a few base pairs. Single base-pair substi-
tutions can create missense codons or nonsense (tran-

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Solved problems 477

pyrimidine (or vice versa) is called a transversion. Single- Cellular enzymes participate in mutation in several
base-pair additions or deletions (indels) produce frame- ways. Errors of enzymes can result in altered DNA. Repli-
shift mutations. Certain human genes that contain tri- cation is often needed to fix the new base stably in the
nucleotide repeats — especially those that are expressed DNA. Several enzymes specialize in repair. Some of
in neural tissue — become mutated through the expan- these provide accurate reversals; others are error-prone
sion of these repeats and can thus cause disease. The for- and result in mutations. Repair is by direct reversal of
mation of monoamino acid repeats within the polypep- damage, excision and resynthesis using existing templates,
tides encoded by these genes is responsible for the postreplication repair, transcription-coupled repair, end-
mutant phenotypes. joining, or repair by homologous recombination.

Mutations can be either spontaneous or induced The molecular mechanism of crossing-over is
by mutagenic radiation or chemicals. Spontaneous thought to involve repairlike processes. In the current
changes are generally a range of types. Mutagens often best model, a double-stranded break in one chromatid
result in a specific type of change because of their initiates formation of a pair of single-stranded Holliday
chemical specificity. For example, some produce ex- junctions that can be resolved to result in a one stan-
clusively G · C : A · T transitions; others, exclu- dard double-helical crossover. Heteroduplex DNA is
sively frameshifts. formed in the process, and if this happens to span a
heterozygous gene, aberrant ascus ratios can result. Cer-
Spontaneous base replacement can be from a tain steps in the double-strand break model (such as
type of chemical isomerization called tautomeric shift. double-strand breaks, exonuclease activity, mismatch
Some chemicals exacerbate this type of change. Some repair, polymerase activity, ligase activity) are very simi-
mutagens alter the structure of a base, leading to new lar to those in several types of mutational repair. Com-
hydrogen-bonding properties. Some agents (such as mon enzymes are involved, and mutants observed for
UV) cause large-scale damage to the base or base one system often affect the other.
removal.

KEY TERMS gene conversion (p. 474) nonconservative substitution (p. 456)
heteroduplex DNA (p. 474) nonsense mutations (p. 455)
acridine orange (p. 458) homology-dependent repair nucleotide excision-repair
adenine methylase (p. 471)
aflatoxin B1 (AFB1) (p. 461) systems (p. 469) system (p. 470)
apurinic site (p. 461) hot spots (p. 457) oxidatively damaged bases (p. 464)
base analogs (p. 457) ICR compounds (p. 458) proflavin (p. 458)
base excision repair (p. 469) imino (p. 457) replica plating (p. 462)
conservative substitution (p. 456) indel mutations (p. 465) SOS system (p. 459)
deamination (p. 464) intercalating agents (p. 458) spontaneous lesions (p. 463)
depurination (p. 463) keto (p. 457) synonymous mutations (p. 455)
DNA glycosylases (p. 469) mismatch-repair system (p. 471) tautomeric shift (p. 457)
double-strand break (p. 472) missense mutations (p. 455) tautomers (p. 457)
enol (p. 457) mutagenesis (p. 453) transition (p. 454)
fluctuation test (p. 462) mutational specificity (p. 457) transversion (p. 454)
frameshift mutations (p. 456) trinucleotide repeat (p. 466)

SOLVED PROBLEMS Solution

1. In Chapter 9, we learned that UAG and UAA EMS induces primarily G и C : A и T transitions. UAG
codons are two of the chain-terminating nonsense codons could not be reverted to wild type, because only
triplets. On the basis of the specificity of aflatoxin the UAG : UAA change would be stimulated by EMS
B1 and ethylmethanesulfonate (EMS), describe and that generates a nonsense (ochre) codon. UAA codons
whether each mutagen would be able to revert would not be acted on by EMS. Aflatoxin B1 induces
these codons to wild type.

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478 Chapter 14 • Mutation, Repair, and Recombination

primarily G и C : T и A transversions. Only the third by causing the substitution of one base for another. This
position of UAG codons would be acted on, resulting in a substitution cannot compensate for the frameshift re-
UAG : UAU change (on the mRNA level), which pro- sulting from ICR-191 and acridines.
duces tyrosine. Therefore, if tyrosine were an acceptable
amino acid at the corresponding site in the protein, 3. A mutant of E. coli is highly resistant to mutagenesis
aflatoxin B1 could revert UAG codons. Aflatoxin B1 by a variety of agents, including ultraviolet light,
would not revert UAA codons, because no G и C base aflatoxin B1, and benzo(a)pyrene. Explain one possi-
pairs appear at the corresponding position in the DNA. ble cause of this mutant phenotype.

2. Explain why mutations induced by acridines in Solution
phage T4 or by ICR-191 in bacteria cannot be re-
verted by 5-bromouracil. The mutant might lack the SOS system and perhaps
carries a defect in the UmuC gene. Such strains would
Solution not be able to bypass replication-blocking lesions of the
type caused by the three mutagens listed. Without the
Acridines and ICR-191 induce mutations by deleting or processing of premutational lesions, mutations would
adding one or more base pairs, which results in a not be recovered in viable cells.
frameshift. However, 5-bromouracil induces mutations

PROBLEMS 8. a. Acridine orange is an effective mutagen for pro-
ducing null alleles by mutation. Why do you think
BASIC PROBLEMS this is so?
1. Consider the wild-type and mutant sequences below:
b. A certain acridinelike compound generates only
Wild ....CTTGCAAGCGAATC.... single insertions. A mutation induced with this com-
Mutant ....CTTGCTAGCGAATC.... pound is treated with the same compound, and
some revertants are produced. How is this possible?
The substitution shown seems to have created a stop
codon. What further information do you need to be 9. A newly discovered SOS bypass system is found to
confident that this is so? preferentially insert thymine opposite apurinic sites.
What type of mutations should be preferentially
2. What type of mutation is the following (shown as produced?
mRNA)?
10. Draw diagrams that contrast replication slippage and
Wild type .....5ЈAAUCCUUACGGA 3Ј..... asymmetrical crossovers as possible causes of multi-
Mutant .....5ЈAAUCCUACGGA 3Ј....... ple tandem repeats.

3. Can a missense mutation of proline to histidine be 11. In a project in which she is trying to induce muta-
made with a G и C : A и T transition-causing tions using UV radiation, a student notices that on
mutagen? What about a proline-to-serine missense bright sunny days far fewer mutations are obtained.
mutation? Suggest an explanation.

4. By base-pair substitution, what are all the synony- 12. A mutational lesion results in a sequence containing
mous changes that can be made starting with the a mismatched base pair:
codon CGG?
5ЈAGCTGCCTT 3Ј
5. a. What are all the transversions that can be made 3ЈACGATGGAA 5Ј
starting with the codon CGG?
Codon
b. Which of these will be missense? Can you be sure?
If mismatch repair occurs in either direction, which
6. Which tautomer of thymine can form the most hy- amino acids could be found at this site?
drogen bonds in pairing with other DNA bases?
13. What aspect of the double-strand break model is re-
7. a. If the enol form of thymine is inserted on a sponsible for the finding that gene conversion is of-
single-stranded template during replication, what ten accompanied by a crossover?
base-pair substitution will result?
b. If thymine enolizes while acting as a template dur- 14. Normally, 6 : 2 aberrant asci are more frequent than
ing replication, what base-pair substitution will result? 5:3 aberrant asci. What might be the explanation for
this on the basis of the double-strand break model?

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Problems 479

15. In the double-strand break model, list all the stages will take place. The following results were obtained
at which exo- and endonucleases act. (a plus sign indicates reversion; HA causes only
G и C : A·T transitions).
16. Differentiate between the elements of the following
pairs: Spontaneous
a. Transitions and transversions
b. Synonymous and neutral mutations Mutant 5-BU HA Proflavin reversion
c. Missense and nonsense mutations
d. Frameshift and nonsense mutations 1 ϪϪ Ϫ Ϫ
2 ϪϪ ϩ ϩ
17. Why are frameshift mutations more likely than mis- 3 ϩϪ Ϫ ϩ
sense mutations to result in proteins that lack nor- 4 ϪϪ Ϫ ϩ
mal function? 5 ϩϩ Ϫ ϩ

18. Diagram two different mechanisms for deletion for- a. For each of the five mutants, describe the nature
mation. What type of information provided from of the original mutation event (not the reversion) at
DNA sequencing can distinguish between these the molecular level. Be as specific as possible.
possibilities?
b. For each of the five mutants, name a possible muta-
19. Describe two spontaneous lesions that can lead to gen that could have caused the original mutation event.
mutations. (Spontaneous mutation is not an acceptable answer.)

20. Compare the mechanism of action of 5-bromouracil c. In the reversion experiment for mutant 5, a par-
(5-BU) with ethylmethanesulfonate (EMS) in caus- ticularly interesting prototrophic derivative is ob-
ing mutations. Explain the specificity of mutagenesis tained. When this type is crossed with a standard
for each agent in light of the proposed mechanism. wild-type strain, the progeny consist of 90 percent
prototrophs and 10 percent auxotrophs. Give a full
21. Compare the two different systems required for the explanation for these results, including a precise rea-
repair of AP sites and the removal of photodimers. son for the frequencies observed.

22. In adult cells that have stopped dividing, what types 28. You are using nitrosoguanidine to “revert” mutant
of repair systems are possible? nic-2 (nicotinamide-requiring) alleles in Neurospora.
www. You treat cells, plate them on a medium without
23. A certain compound that is an analog of the base nicotinamide, and look for prototrophic colonies. You
cytosine can become incorporated into DNA. It nor- obtain the following results for two mutant alleles.
mally hydrogen-bonds just as cytosine does, but it Explain these results at the molecular level, and
quite often isomerizes to a form that hydrogen- indicate how you would test your hypotheses.
bonds as thymine does. Do you expect this com-
pound to be mutagenic, and if so, what types of a. With nic-2 allele 1, you obtain no prototrophs at all.
changes might it induce at the DNA level?
b. With nic-2 allele 2, you obtain three prototrophic
24. Describe the repair systems that operate after colonies A, B, and C, and you cross each separately
depurination and deamination. with a wild-type strain. From the cross prototroph
A ϫ wild type, you obtain 100 progeny, all of which
25. Describe the model for indel formation. Show how are prototrophic. From the cross prototroph
this model can explain mutational hot spots in the B ϫ wild type, you obtain 100 progeny, of which 78
lacI gene of E. coli. are prototrophic and 22 are nicotinamide-requiring.
From the cross prototroph C ϫ wild type, you ob-
CHALLENGING PROBLEMS tain 1000 progeny, of which 996 are prototrophic and
4 are nicotinamide-requiring.
26. a. Why is it impossible to induce nonsense muta-
tions (represented at the mRNA level by the triplets 29. Fill in the following table, using a plus sign (ϩ) to
UAG, UAA, and UGA) by treating wild-type strains indicate that the mutagenic lesion (base damage) in-
with mutagens that cause only A и T : G и C tran- duces the indicated base change and a minus sign
sitions in DNA? (Ϫ) to show that it does not.
b. Hydroxylamine (HA) causes only G и C : A и T
transitions in DNA. Will HA produce nonsense mu- Base change O-6-Methyl 8-Oxo C – C
tations in wild-type strains? G dG photodimer
c. Will HA treatment revert nonsense mutations?
A и T to G и T
27. Several auxotrophic point mutants in Neurospora G и C to T и A
are treated with various agents to see if reversion G и C to A и T

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480 Chapter 14 • Mutation, Repair, and Recombination

30. You are working with a newly discovered mutagen, (7) general nucleotide excision repair
and you wish to determine the base change that it
introduces into DNA. Thus far you have determined (8) methyl-directed mismatch repair
that the mutagen chemically alters a single base in a. Deamination of cytosine
such a way that its base-pairing properties are al- b. 8-Oxo dG
tered permanently. In order to determine the speci- c. Aflatoxin B1 adduct
ficity of the alteration, you examine the amino acid d. G и T mispair as replication error
changes that take place after mutagenesis. A sample e. 5Ј-CC-3Ј dimer
of what you find is shown below: f. AP site
g. O6-methylguanine
Original: Gln-His-Ile-Glu-Lys
Mutant: Gln-His-Met-Glu-Lys 34. Which of the following linear asci shows gene con-
version at the arg-2 locus?
Original: Ala-Val-Asn-Arg
Mutant: Ala-Val-Ser-Arg 123456
111111
Original: Arg-Ser-Leu 111111
Mutant: Arg-Ser-Leu-Trp-Lys-Thr-Phe 1 arg 1 1 arg arg
1 arg arg arg arg arg
What is the base-change specificity of the mutagen? arg arg arg 1 1 arg
arg arg arg arg 1 arg
31. You now find an additional mutant from the experi- arg 1 arg arg arg arg
ment in Problem 30: arg 1 arg arg arg arg

Original: Ile-Leu-His-Gln
Mutant: Ile-Pro-His-Gln

Could the base-change specificity in your answer 35. Assume you have made a cross in Neurospora using
to Problem 30 account for this mutation? Why or a mutant that has three mutant sites in the same
why not? gene, called 1, 2 and 3, which are spaced evenly
through the 2-kb gene:
32. Strain A of Neurospora contains an ad-3 mutation
that reverts spontaneously at a rate of 1026. Strain A 123 ϫ 111
is crossed with a newly acquired wild-type isolate,
and ad-3 strains are recovered from the progeny. Explain a likely origin of the following two asci:
When 28 different ad-3 progeny strains are exam-
ined, 13 lines are found to revert at the rate of 1026, 123 123
but the remaining 15 lines revert at the rate of 1023. 123 123
Formulate a hypothesis to account for these find- 111 121
ings, and outline an experimental program to test 111 121
your hypothesis. 111 111
111
33. For each lesion in parts a – g, indicate which of the 111 111
following repair systems repairs that lesion: 111
111

111

(1) alkyltransferase 36. The double-strand break model illustrated in this
chapter generated one heteroduplex. Other models
(2) endonuclease generate two identical heteroduplexes in the same
meiosis, so that the chromatids are parent 1, hetero-
(3) photolyase duplex, heteroduplex, parent 2. What octad patterns
would be produced from the various combinations
(4) MutY glycosylase of repair and nonrepair of these two heteroduplex
mismatches?
(5) MutM glycosylase

(6) uracil DNA glycosylase

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15

LARGE-SCALE
CHROMOSOMAL CHANGES

A reciprocal translocation demonstrated by chromosome KEY QUESTIONS
painting. A suspension of chromosomes from many cells is
passed through an electronic device that sorts them by size. • How common are polyploids (organisms
DNA is extracted from individual chromosomes, denatured, with multiple chromosome sets)?
joined with one of several fluorescent dyes, and then added to
partially denatured chromosomes on a slide. The fluorescent • How do polyploids arise?
DNA “finds” its own chromosome and binds along its length
by base complementarity, thus “painting” it. In this • Do polyploids have any special properties?
preparation, a bright blue and a pink dye have been used to
paint different chromosomes. The preparation shows one • Is the polyploid state transmissible
normal pink chromosome, one normal light blue, and two that to offspring?
have exchanged their tips. [Lawrence Berkeley Laboratory.]
• What inheritance patterns are observed
in the progeny of polyploids?

• How do aneuploids (variants in which
a single chromosome has been gained
or lost) arise?

• Do aneuploids show any special properties?

• What inheritance patterns are produced
by aneuploids?

• How do large-scale chromosome
rearrangements (deletions, duplications,
inversions, and translocations) arise?

• Do these rearrangements have any
special properties?

• What inheritance patterns are produced
by rearrangements?

OUTLINE

15.1 Changes in chromosome number
15.2 Changes in chromosome structure
15.3 Overall incidence of human

chromosome mutations

481

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482 Chapter 15 • Large-Scale Chromosomal Changes

CHAPTER OVERVIEW Figure 15-1 Children with Down syndrome.

A young couple is planning to have children. The hus- [Bob Daemmrich/The Image Works.]
band knows that his grandmother had a child with
Down syndrome by a second marriage. Down syndrome chromosome mutations are changes to a chromosome re-
is a set of physical and mental disorders caused by the gion encompassing multiple genes. Chromosome muta-
presence of an extra chromosome 21 (Figure 15-1). No tions can be detected by microscopic examination, ge-
records of the birth, which occurred early in the twenti- netic analysis, or both. In contrast, gene mutations are
eth century, are available, but the couple knows of no never detectable microscopically; a chromosome bearing
other cases of Down syndrome in their families. a gene mutation looks the same under the microscope as
one carrying the wild-type allele. Chromosome mutations
The couple has heard that Down syndrome results
from a rare chance mistake in egg production and there-
fore decide that they stand only a low chance of having
such a child. They decide to have children. Their first
child is unaffected, but the next conception aborts spon-
taneously (a miscarriage), and their second child is born
with Down syndrome. Was this a coincidence, or is it
possible that there is a connection between the genetic
makeup of the man and that of his grandmother that led
to their both having Down syndrome children? Was the
spontaneous abortion significant? What tests might be
necessary to investigate this situation? The analysis of
such questions is the topic of this chapter.

We saw in Chapter 10 that gene mutations are one
source of genomic change. However, the genome can also
be remodeled on a larger scale by alterations to chromo-
some structure or by changes in the number of copies of
chromosomes in a cell. These large-scale variations are
termed chromosome mutations to distinguish them from
gene mutations. Broadly speaking, gene mutations are de-
fined as changes that take place within a gene, whereas

CHAPTER OVERVIEW Figure

Relocation of Deletion Loss of
genetic material genetic material
Missing chromosome(s)
Translocation
Extra chromosome(s)
From another Wild-type sequence
chromosome

Inversion

Duplication Gain of
genetic material

Figure 15-2 Overview of chromosome mutations. The figure has been divided into three
colored regions to depict the main types of chromosome mutations that can occur. These
involve the loss, gain, or relocation of entire chromosomes or chromosome segments.
The wild-type chromosome is shown in the center.

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15.1 Changes in chromosome number 483

have been best characterized in eukaryotes, and all the ex- of chromosomes present in our cells. Foremost is the fact
amples in this chapter are from that group. that a group of common genetic disorders results from
the presence of an abnormal number of chromosomes.
Chromosome mutations are important from several Although this group of disorders is small, it accounts for
biological perspectives. First, they can be sources of insight a large proportion of the genetically determined health
into how genes act in concert on a genomic scale. Second, problems that afflict humans. Also of relevance to hu-
they reveal several important features of meiosis and chro- mans is the role of chromosome mutations in plant
mosome architecture. Third, they constitute useful tools breeding: plant breeders have routinely manipulated
for experimental genomic manipulation. Fourth, they are chromosome number to improve commercially impor-
sources of insight into evolutionary processes. tant agricultural crops.

Many chromosome mutations cause abnormalities Changes in chromosome number are of two basic
in cell and organismal function. Most of these abnormal- types: changes in whole chromosome sets, resulting in a
ities stem from changes in gene number or gene position. condition called aberrant euploidy, and changes in parts
In some cases, a chromosome mutation results from of chromosome sets, resulting in a condition called
chromosome breakage. If the break occurs within a gene, aneuploidy.
the result is functional disruption of that gene.
Aberrant euploidy
For our purposes, we shall divide chromosome muta-
tions into two groups: changes in chromosome number Organisms with multiples of the basic chromosome set
and changes in chromosome structure. These two groups (genome) are referred to as euploid. We learned in ear-
represent two fundamentally different kinds of events. lier chapters that familiar eukaryotes such as plants, ani-
Changes in chromosome number are not associated with mals, and fungi carry in their cells either one chromo-
structural alterations of any of the DNA molecules of the some set (haploidy) or two chromosome sets (diploidy).
cell. Rather, it is the number of these DNA molecules that In these species, the haploid and diploid states are both
is changed, and this change in number is the basis of their cases of normal euploidy. Organisms that have more or
genetic effects. Changes in chromosome structure, on the fewer than the normal number of sets are aberrant eu-
other hand, result in novel sequence arrangements within ploids. Polyploids are individual organisms that have
one or more DNA double helices. These two types of more than two chromosome sets. They can be repre-
chromosome mutations are illustrated in Figure 15-2, sented by 3n (triploid), 4n (tetraploid), 5n (penta-
which is a summary of the topics of this chapter. We be- ploid), 6n (hexaploid), and so forth. (The number of
gin by exploring the nature and consequences of changes chromosome sets is called the ploidy or ploidy level.) An
in chromosome number. individual of a normally diploid species that has only
one chromosome set (n) is called a monoploid to distin-
15.1 Changes in guish it from an individual of a normally haploid species
chromosome number (also n). Examples of these conditions are shown in the
first four rows of Table 15-1.
In genetics as a whole, few topics impinge on human af-
fairs quite so directly as that of changes in the number

TABLE 15-1 Chromosome Constitutions in a Normally Diploid Organism
with Three Chromosomes (Labeled A, B, and C) in the Basic Set

Name Designation Constitution Number of
chromosomes
Euploids n ABC
Monoploid 2n AA BB CC 3
Diploid 3n AAA BBB CCC 6
Triploid 4n AAAA BBBB CCCC 9
Tetraploid 12
2n Ϫ 1 A BB CC
Aneuploids AA B CC 5
Monosomic 2n ϩ 1 AA BB C 5
AAA BB CC 5
Trisomic AA BBB CC 7
AA BB CCC 7
7

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484 Chapter 15 • Large-Scale Chromosomal Changes

MONOPLOIDS Male bees, wasps, and ants are mono- been an important factor in the origin of new plant species.
ploid. In the normal life cycles of these insects, males de- The evidence for this is shown in Figure 15-3, which dis-
velop by parthenogenesis (the development of a special- plays the frequency distribution of haploid chromosome
ized type of unfertilized egg into an embryo without the numbers in dicotyledonous plant species. Above a haploid
need for fertilization). In most other species, however, number of about 12, even numbers are much more com-
monoploid zygotes fail to develop. The reason is that vir- mon than odd numbers. This pattern is a consequence of
tually all individuals in a diploid species carry a number the polyploid origin of many plant species, because dou-
of deleterious recessive mutations, together called a bling and redoubling of a number can give rise only to even
“genetic load.” The deleterious recessive alleles are numbers. Animal species do not show such a distribution,
masked by wild-type alleles in the diploid condition, but owing to the relative rareness of polyploid animals.
are automatically expressed in a monoploid derived
from a diploid. Monoploids that do develop to advanced In aberrant euploids, there is often a correlation be-
stages are abnormal. If they survive to adulthood, their tween the number of copies of the chromosome set and
germ cells cannot proceed through meiosis normally be- the size of the organism. A tetraploid organism, for ex-
cause the chromosomes have no pairing partners. Thus, ample, typically looks very similar to its diploid counter-
monoploids are characteristically sterile. (Male bees, part in its proportions, except that the tetraploid is big-
wasps, and ants bypass meiosis; in these groups, gametes ger, both as a whole and in its component parts. The
are produced by mitosis.) higher the ploidy level, the larger the size of the organ-
ism (Figure 15-4).
POLYPLOIDS Polyploidy is very common in plants but
rarer in animals (for reasons that we will consider later). In- MESSAGE Polyploid plants are often larger and have
deed, an increase in the number of chromosome sets has larger component parts than their diploid relatives.

1100 n=8
12

1000

900

800 14
700

Number of species 600
500 16 18

400

20 24
300 22

200 28
26

100 30 32 34 36
38 40

0 1 5 10 20 30 40 50 60

Haploid number

Figure 15-3 Frequency distribution of haploid chromosome number in dicot plants. Notice the
excess of even-numbered values in the higher ranges, suggesting ancestral polyploidization.

[Adapted from Verne Grant, The Origin of Adaptations. Columbia University Press, 1963.]

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15.1 Changes in chromosome number 485

2n are only homeologous (partially homologous), not fully
homologous as they are in autopolyploids.
Stoma
Autopolyploids Triploids are usually autopolyploids.
(a) They arise spontaneously in nature, but they can be
constructed by geneticists from the cross of a 4n
4n (tetraploid) and a 2n (diploid). The 2n and the n ga-
metes produced by the tetraploid and the diploid, re-
(b) spectively, unite to form a 3n triploid. Triploids are
characteristically sterile. The problem, as in monoploids,
8n lies in pairing at meiosis. The molecular mechanisms for
synapsis, or true pairing, dictate that pairing can take
(c) place between only two of the three chromosomes of
each type (Figure 15-5). Paired homologs (bivalents)
Figure 15-4 Epidermal leaf cells of tobacco plants with segregate to opposite poles, but the unpaired homologs
increasing ploidy. Cell size increases, particularly evident (univalents) pass to either pole randomly. In the case of
in stoma size, with an increase in ploidy. (a) Diploid; a trivalent, a paired group of three, the paired cen-
(b) tetraploid; (c) octoploid. [From W. Williams, Genetic Principles tromeres segregate as a bivalent and the unpaired one as
a univalent. These segregations take place for every chro-
and Plant Breeding. Blackwell Scientific Publications, Ltd.] mosome threesome, so for any chromosomal type, the
gamete could receive either one or two chromosomes. It
In the realm of polyploids, we must distinguish is unlikely that a gamete will receive two for every chro-
between autopolyploids, which have multiple chromo- mosomal type, or that it will receive one for every
some sets originating from within one species, and allo- chromosomal type. Hence the likelihood is that gametes
polyploids, which have sets from two or more different will have chromosome numbers intermediate between
species. Allopolyploids form only between closely re- the haploid and diploid number; such genomes are of a
lated species; however, the different chromosome sets type called aneuploid (“not euploid”).

Aneuploid gametes generally do not give rise to vi-
able offspring. In plants, aneuploid pollen grains are gen-
erally inviable and hence unable to fertilize the female ga-
mete. In any organism zygotes that might arise from the
fusion of a haploid and an aneuploid gamete will them-
selves be aneuploid, and typically these zygotes also are
inviable. We will examine the underlying reason for the
inviability of aneuploids when we consider gene balance
later in the chapter.

MESSAGE Polyploids with odd numbers of chromosome
sets, such as triploids, are sterile or highly infertile
because their gametes and offspring are aneuploid.

Pairing
possibilities

or Figure 15-5 Pairing of three
homologous chromosomes. The
Trivalent Bivalent + Univalent three homologous chromosomes of
a triploid may pair in two ways at
meiosis, as a trivalent or as a
bivalent plus a univalent.

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486 Chapter 15 • Large-Scale Chromosomal Changes

Mitosis in a diploid, 2n = 4 Normal Two diploid
cells

With colchicine One tetraploid
cell
Figure 15-6 The use of colchicine to generate a tetraploid from a diploid. Colchicine 4n = 8
added to mitotic cells during metaphase and anaphase disrupts spindle-fiber formation,
preventing the migration of chromatids after the centromere is split. A single cell is
created that contains pairs of identical chromosomes that are homozygous at all loci.

Autotetraploids arise by the doubling of a 2n com- chromosome segregation or cell division does not. As the
plement to 4n. This doubling can occur spontaneously, treated cell enters telophase, a nuclear membrane forms
but it can also be induced artificially by applying chemi- around the entire doubled set of chromosomes. Thus,
cal agents that disrupt microtubule polymerization. We treating diploid (2n) cells with colchicine for one cell
saw in Chapter 3 that chromosome segregation is pow- cycle leads to tetraploids (4n) with exactly four copies
ered by spindle fibers, which are polymers of the protein of each type of chromosome (Figure 15-6). Treatment
tubulin. Hence disruption of microtubule polymeriza- for an additional cell cycle produces octoploids (8n), and
tion blocks chromosome segregation. The chemical so forth. This method works in both plant and animal
treatment is normally applied to somatic tissue during cells, but generally plants seem to be much more toler-
the formation of spindle fibers in cells undergoing divi- ant of polyploidy. Note that all alleles in the genotype
sion. The resulting polyploid tissue (such as a polyploid are doubled. Therefore, if a diploid cell of genotype
branch of a plant) can be detected by examining stained A/a ; B/b is doubled, the resulting autotetraploid will be
chromosomes from the tissue under a microscope. Such of genotype A/A/a/a ; B/B/b/b.
a branch can be removed and used as a cutting to gener-
ate a polyploid plant or allowed to produce flowers, Because 4 is an even number, autotetraploids can
which when selfed would produce polyploid offspring. have a regular meiosis, although this is by no means
A commonly used antitubulin agent is colchicine, an al- always the case. The crucial factor is how the four
kaloid extracted from the autumn crocus. In colchicine- chromosomes of each set pair and segregate. There are
treated cells, the S phase of the cell cycle occurs, but several possibilities, as shown in Figure 15-7. If the chro-
mosomes pair as bivalents or quadrivalents, the chromo-

www. ANIMATED ART Autotetraploid meiosis Pairing
possibilities

Two bivalents One quadrivalent Univalent + Trivalent Figure 15-7 Three different
meiotic pairing possibilities in
tetraploids. The four homologous
chromosomes may pair as two
bivalents or a quadrivalent, and
each can yield functional
gametes. A third possibility, a
trivalent plus a univalent, yields
nonfunctional gametes.

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15.1 Changes in chromosome number 487

somes segregate normally, producing diploid gametes. to allow intercrossing. Fusion of an n1 and an n2 gamete
produced a viable hybrid progeny individual of constitu-
The fusion of gametes at fertilization regenerates the tion n1 ϩ n2 ϭ 18. However, this hybrid was functionally
sterile because the 9 chromosomes from the cabbage
tetraploid state. If trivalents form, segregation leads to parent were different enough from the radish chromo-
somes that pairs did not synapse and segregate normally
nonfunctional aneuploid gametes, and hence sterility. at meiosis, and thus the hybrid could not produce func-
tional gametes.
What genetic ratios are produced by an autote-
Eventually, one part of the hybrid plant produced
traploid? Assume for simplicity that the tetraploid forms some seeds. On planting, these seeds produced fertile in-
dividuals with 36 chromosomes. All these individuals
only bivalents. If we start with an A/A/a/a tetraploid were allopolyploids. They had apparently been derived
from spontaneous, accidental chromosome doubling to
plant and self it, what proportion of progeny would be 2n1 ϩ 2n2 in one region of the sterile hybrid, presum-
ably in tissue that eventually became a flower and un-
a/a/a/a? Obviously we first need to deduce the fre- derwent meiosis to produce gametes. In 2n1 ϩ 2n2 tis-
sue, there is a pairing partner for each chromosome, and
quency of a/a gametes because this is the only type that functional gametes of the type n1 ϩ n2 are produced.
These gametes fuse to give 2n1 ϩ 2n2 allopolyploid
can produce a recessive homozygote. The a/a gametes progeny, which also are fertile. This kind of allopolyploid
is sometimes called an amphidiploid, or doubled diploid
can arise only if the pairings are both A with a, and then (Figure 15-8). Treating a sterile hybrid with colchicine
greatly increases the chances that the chromosome sets
the a alleles must both segregate to the same pole. Let’s will double. Amphidiploids are now synthesized rou-
tinely in this manner. (Unfortunately for Karpechenko,
calculate the frequencies of the possible outcomes by his amphidiploid had the roots of a cabbage and the
leaves of a radish.)
means of the following thought experiment. Consider
When Karpechenko’s allopolyploid was crossed with
the options from the point of view of one of the a chro- either parental species — the cabbage or the radish —
sterile offspring resulted. The offspring of the cross with
mosomes faced with the options of pairing with the cabbage were 2n1 ϩ n2, constituted from an n1 ϩ n2

other a chromosome or with one of the two A chromo-

somes; if pairing is random there is a 2 chance that it will
3
pair with an A chromosome. If it does, then the pairing

of the remaining two chromosomes will necessarily also

be A with a because those are the only chromosomes re-

maining. With these two A-with-a pairings there are two

equally likely segregations, and overall 1 of the products
4
will contain both a alleles at one pole. Hence the proba-

bility of an a/a gamete will be 2 ϫ 1 ϭ 61. Hence, if
3 4
gametes pair randomly, the probability of an a/a/a/a

zygote will be 1 ϫ 1 ϭ 316, and by subtraction the prob-
6 6 be 3365. Therefore a 35 : 1 pheno-
ability of A/ – / – / – will

typic ratio is expected.

Genomic sequencing has shown that many species

that act like “normal” diploids or haploids are in fact de-

scendants of autopolyploids that occurred in past evolu-

tionary times. For example, genomic analysis of the hap- n=9
Gametes
loid yeast Saccharomyces cerevisiae has shown that most
n=9
chromosomal regions have a duplicate somewhere else

in the genome. In fact the ancestral genome of this yeast doSupblionntganeous

was probably very similar to that of the filamentous fun-

gus Ashbya gossypii, as revealed by comparisons of the Raphanus
2n = 18
size and gene content of their fully sequenced genomes.
Parents ×
Hence the smaller Ashbya genome presumably doubled
Brassica
at some point in the distant past. Subsequent rearrange- 2n = 18 Sterile F1 hybrid
n+n=9+9
ment, mutation, and partial loss of segments gave rise to Raphanobrassica
2n = 18
the modern yeast species.

Fertile amphidiploid

Allopolyploids An allopolyploid is a plant that is a hybrid 2n + 2n = 18 + 18
of two or more species, containing two or more copies of
each of the input genomes. The prototypic allopolyploid 4n = 36
was an allotetraploid synthesized by G. Karpechenko in
1928. He wanted to make a fertile hybrid that would Figure 15-8 The origin of the amphidiploid (Raphanobrassica)
have the leaves of the cabbage (Brassica) and the roots from cabbage (Brassica) and radish (Raphanus). The fertile
of the radish (Raphanus), because these were the agri- amphidiploid arose from spontaneous doubling in the 2n = 18
culturally important parts of each plant. Each of these sterile hybrid. [From A. M. Srb, R. D. Owen, and R. S. Edgar,
two species has 18 chromosomes, so 2n1 ϭ 2n2 ϭ 18,
and n1 ϭ n2 ϭ 9. The species are related closely enough General Genetics, 2d ed. Copyright 1965 by W. H. Freeman. Adapted

from G. Karpechenko, Z. Indukt. Abst. Vererb. 48, 1928, 27.]

44200_15_p481-520 3/22/04 4:07 PM Page 488 Chapter 15 • Large-Scale Chromosomal Changes

488

B. oleracea,
2n = 18

Cabbage
Cauliflower

Broccoli
Kale

Kohlrabi
Brussels sprouts

n=9 n=9

B. carinata, B. napus,
2n = 34 2n = 38

Abyssinian mustard Rutabaga
Oil rape

n=8 n = 10 Figure 15-9
Three species of
B. nigra, n=8 B. juncea, n = 10 B. campestris, Brassica (blue
2n = 16 2n = 36 2n = 20 boxes) and their
allopolyploids (pink
Black mustard Leaf mustard Chinese cabbage boxes), showing the
Turnip importance of
allopolyploidy in the
Turnip rape production of new
species.

gamete from the allopolyploid and an n1 gamete from than previously thought, perhaps contributing to its
the cabbage. The n2 chromosomes had no pairing part- potential for adaptation.
ners; hence, a normal meiosis could not take place, and
the offspring were sterile. Thus, Karpechenko had effec- A particularly interesting natural allopolyploid is
tively created a new species, with no possibility of gene bread wheat, Triticum aestivum (6n ϭ 42). By studying its
exchange with either cabbage or radish. He called his wild relatives, geneticists have reconstructed a probable
new plant Raphanobrassica. evolutionary history of this plant. Figure 15-10 shows
that bread wheat is composed of two sets each of three
In nature, allopolyploidy seems to have been a ma- ancestral genomes. At meiosis, pairing is always between
jor force in the evolution of new plant species. One homologs from the same ancestral genome. Hence, in
convincing example is shown by the genus Brassica, as bread wheat meiosis, there are always 21 bivalents.
illustrated in Figure 15-9. Here three different parent
species have hybridized in all possible pair combina- Allopolyploid plant cells can also be produced artifi-
tions to form new amphidiploid species. Natural poly- cially by fusing diploid cells from different species. First
ploidy was once viewed as a somewhat rare occurrence, the walls of two diploid cells are removed by treatment
but recent work has shown that it is a recurrent event with an enzyme, and the membranes of the two cells
in many plant species. The use of DNA markers has fuse and become one. The nuclei often fuse, too, result-
made it possible to show that polyploids in any popula- ing in the polyploid. If the cell is nurtured with the ap-
tion or area which appear to be the same can have propriate hormones and nutrients, it divides to become a
many different parental genotypes as a result of many small allopolyploid plantlet, which can then be trans-
independent past fusions. It has been estimated that ferred to soil.
about 50 percent of all angiosperm plants are poly-
ploids, resulting from auto- or allopolyploidy. As a re- MESSAGE Allopolyploid plants can be synthesized by
sult of multiple polyploidizations, the amount of allelic crossing related species and doubling the chromosomes of
variation within a polyploid species is much higher the hybrid or by fusing diploid cells.

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15.1 Changes in chromosome number 489

AGRICULTURAL APPLICATIONS
Variations in chromosome number
have been exploited to create new
plant lines with desirable features.
Some examples follow.

Cultivated as A wild diploid wheat × A wild diploid wheat Monoploids Diploidy is an inherent
Einkorn wheat T. monococcum possibly T. searsii nuisance for plant breeders. When
they want to induce and select new
AA AA BB

recessive mutations that are favor-

A B able for agricultural purposes, the

new mutations cannot be detected
(× 2) unless they are homozygous. Breed-

ers may also want to find favorable

A wild tetrapoloid wheat new combinations of alleles at dif-
T. turgidum ferent loci, but such favorable allele

AA BB combinations in heterozygotes will

be broken up by recombination at

meiosis. Monoploids provide a way

A wild diploid wheat × Cultivated 10,000 yr around some of these problems.
Monoploids can be artificially
T. tauschii B.P. as Emmer wheat
derived from the products of meio-
DD AA BB sis in the plant’s anthers. A cell des-

A BD tined to become a pollen grain can
instead be induced by cold treat-

(× 2; ~ 8000 yr ago) ment to grow into an embryoid, a
small dividing mass of monoploid

cells. The embryoid can be grown on agar to form a

Hexaploid wheat monoploid plantlet, which can then be potted in soil and

T. aestivum allowed to mature (Figure 15-11).
AA BB DD Plant monoploids can be exploited in several ways.

In one approach, they are first examined for favorable al-

lelic combinations that have arisen from the recombina-

Figure 15-10 The proposed tion of alleles already present in a heterozygous diploid
evolutionary history of modern parent. Hence from a parent that is A/a ; B/b might
hexaploid wheat. Amphidiploids come a favorable monoploid combination a ; b. The
were produced at two points in monoploid can then be subjected to chromosome dou-
this history. A, B, and D are bling to produce homozygous diploid cells, a/a ; b/b,

different chromosome sets. that are capable of normal reproduction.

Anthers

Diploid Immature pollen Monoploid Monoploid Monoploid Figure 15-11 Generating
plant cells plated embryoids grow plantlet plant monoploid plants by tissue
culture.

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490 Chapter 15 • Large-Scale Chromosomal Changes

Another approach is to treat monoploid cells Allopolyploids Allopolyploidy (formation of polyploids
basically as a population of haploid organisms in a between different species) has been important in the
mutagenesis-and-selection procedure. A population of production of modern crop plants. New World cotton is
monoploid cells is isolated, their walls are removed by a natural allopolyploid that occurred spontaneously, as is
enzymatic treatment, and they are exposed to a muta- wheat. Allopolyploids also are synthesized artificially to
gen. They are then plated on a medium that selects for combine the useful features of parental species into one
some desirable phenotype. This approach has been used type. Only one synthetic amphidiploid has ever been
to select for resistance to toxic compounds produced by widely used commercially, a crop known as Triticale.
a plant parasite as well as to select for resistance to her- This is an amphidiploid between wheat (Triticum, 6n ϭ
bicides being used by farmers to kill weeds. Resistant 42) and rye (Secale, 2n ϭ 14). Hence, for Triticale, 2n ϭ
plantlets eventually grow into monoploid plants, whose 2 ϫ (21 ϩ 7) ϭ 56. This novel plant combines the high
chromosome number can then be doubled using yields of wheat with the ruggedness of rye.
colchicine. This treatment produces diploid tissue and
eventually, by taking a cutting or by selfing a flower, a POLYPLOID ANIMALS Polyploidy is more common in
fully resistant diploid plant. These powerful techniques plants than in animals, but there are cases of naturally
can circumvent the normally slow process of meiosis- occurring polyploid animals. Polyploid species of flat-
based plant breeding. They have been successfully ap- worms, leeches, and brine shrimps reproduce by
plied to important crop plants such as soybeans and parthenogenesis. Triploid and tetraploid Drosophila have
tobacco. been synthesized experimentally. However, examples are
not limited to these so-called lower forms. Naturally oc-
MESSAGE Geneticists can create new plant lines by curring polyploid amphibians and reptiles are surpris-
producing monoploids with favorable genotypes and then ingly common. They have several modes of reproduc-
doubling their chromosomes to form fertile, homozygous tion: polyploid species of frogs and toads participate in
diploids. sexual reproduction, whereas polyploid salamanders and
lizards are parthenogenetic. The Salmonidae (the family
Autotriploids The bananas that are widely available of fishes that includes salmon and trout) provide a famil-
commercially are sterile triploids with 11 chromosomes iar example of the numerous animal species that
in each set (3n ϭ 33). The most obvious expression of appear to have originated through ancestral polyploidy.
the sterility of bananas is the absence of seeds in the
fruit that we eat. (The black specks in bananas are not The sterility of triploids has been commercially ex-
seeds; banana seeds are rock hard — real tooth-breakers.) ploited in animals as well as in plants. Triploid oysters
Seedless watermelons are another example of the com- have been developed because they have a commercial
mercial exploitation of triploidy in plants. advantage over their diploid relatives. The diploids go
through a spawning season, when they are unpalatable,
Autotetraploids Many autotetraploid plants have been but the sterile triploids do not spawn and are palatable
developed as commercial crops to take advantage of year-round.
their increased size (Figure 15-12). Large fruits and
flowers are particularly favored. Aneuploidy

Figure 15-12 Diploid (left) and tetraploid (right) grapes. Aneuploidy is the second major category of chromoso-
mal aberrations in which the chromosome number is
[Copyright Leonard Lessin/Peter Arnold Inc.] abnormal. An aneuploid is an individual organism
whose chromosome number differs from the wild type
by part of a chromosome set. Generally, the aneuploid
chromosome set differs from the wild type by only one
chromosome or by a small number of chromosomes.
An aneuploid can have a chromosome number either
greater or smaller than that of the wild type. Aneuploid
nomenclature (see Table 15-1) is based on the number
of copies of the specific chromosome in the aneuploid
state. For autosomes in diploid organisms, the aneu-
ploid 2n ϩ 1 is trisomic, 2n Ϫ 1 is monosomic, and
2n Ϫ 2 (the Ϫ2 represents the loss of both homologs of
a chromosome) is nullisomic. In haploids, n ϩ 1 is diso-
mic. Special notation is used to describe sex chromo-
some aneuploids because it must deal with the two dif-
ferent chromosomes. The notation merely lists the

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15.1 Changes in chromosome number 491

copies of each sex chromosome, such as XXY, XYY, a monosomic (2n Ϫ 1) zygote is produced. The fusion of
XXX, or XO (the “O” stands for absence of a chromo- an n ϩ 1 and an n gamete yields a trisomic 2n ϩ 1.
some and is included to show that the single X symbol
is not a typographical error). MESSAGE Aneuploid organisms result mainly from
nondisjunction during a parental meiosis.
NONDISJUNCTION The cause of most aneuploidy is
nondisjunction in the course of meiosis or mitosis. Dis- Nondisjunction occurs spontaneously. Like most
junction is another word for the normal segregation of gene mutations, it is an example of a chance failure of a
homologous chromosomes or chromatids to opposite basic cellular process. The precise molecular processes
poles at meiotic or mitotic divisions. Nondisjunction is a that fail are not known, but in experimental systems, the
failure of this process, in which two chromosomes or frequency of nondisjunction can be increased by inter-
chromatids incorrectly go to one pole and none to the ference with microtubule polymerization action,
other. thereby inhibiting normal chromosome movement. It
appears that disjunction is more likely to go awry in
Mitotic nondisjunction during development results meiosis I. This may not be surprising, because normal
in aneuploid sections of the body (aneuploid sectors). anaphase I disjunction requires that the homologous
Meiotic nondisjunction is more commonly encountered. chromosomes of the tetrad remain paired during
It results in aneuploid meiotic products, leading to de- prophase I and metaphase I, and also requires crossovers.
scendants in which the entire organism is aneuploid. In In contrast, proper disjunction at anaphase II or at mito-
meiotic nondisjunction, the chromosomes may fail to sis requires that the centromere split properly but does
disjoin at either the first or the second meiotic division not require chromosome pairing or crossing-over.
(Figure 15-13). Either way, n Ϫ 1 and n ϩ 1 gametes are
produced. If an n Ϫ 1 gamete is fertilized by an n gamete, Crossovers are a necessary component of the normal
disjunction process. Somehow the formation of a
www. ANIMATED ART Meiotic nondisjunction Nondisjunction Second division n+1 chiasma in a chromosome pair helps to hold the tetrad
at first division n+1 together and ensures that the members of a pair will go
n−1 to opposite poles. In most organisms, the amount of
n−1 crossing-over is sufficient to ensure that all tetrads will
have at least one chiasma per meiosis. In Drosophila,
First division Nondisjunction many of the nondisjunctional chromosomes seen in diso-
at second division mic (n ϩ 1) gametes are nonrecombinant, showing that
n+1 they arise from meioses in which there is no crossing-
n−1 over on that chromosome. Similar observations have
n been made in human trisomies. In addition, in several
different experimental organisms, mutations that inter-
fere with recombination have the effect of massively
increasing the frequency of meiosis I nondisjunction.
All these observations provide evidence for the role of
crossing-over in maintaining chromosome associations in
the tetrad; in the absence of these associations, chromo-
somes are vulnerable to anaphase I nondisjunction.

MESSAGE Crossovers are needed to maintain the intact
tetrad until anaphase I. If crossing-over fails for some
reason, first-division nondisjunction occurs.

n MONOSOMICS (2n ؊ 1) Monosomics are missing one
copy of a chromosome. In most diploid organisms, the
Figure 15-13 The origin of aneuploid gametes by non- absence of one chromosome copy from a pair is delete-
disjunction at the first or second meiotic division. Note that all rious. In humans, monosomics for any of the autosomes
other chromosomes are present in normal number, including die in utero. Many X chromosome monosomics also die
cells in which no chromosomes are shown. in utero, but some are viable. A human chromosome
complement of 44 autosomes plus a single X produces
a condition known as Turner syndrome, represented as

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492 Chapter 15 • Large-Scale Chromosomal Changes

XO. Affected persons have a characteristic phenotype: Short stature Characteristic
they are sterile females, short in stature, and often have facial features
a web of skin extending between the neck and shoul- Low hairline
ders (Figure 15-14). Although their intelligence is near Fold of skin
normal, some of their specific cognitive functions are Shield-shaped Constriction
defective. About 1 in 5000 female births show Turner thorax of aorta
syndrome. Widely spaced Poor breast
nipples development
Geneticists have used viable plant monosomics to Shortened
match newly discovered recessive mutant alleles to a metacarpal IV Elbow
specific chromosome. For example, one can make a set Small deformity
of monosomic lines, each known to lack a different chro- fingernails
mosome. Homozygotes for the new mutant allele are Rudimentary
crossed with each monosomic line, and the progeny of Brown spots (nevi) ovaries
each cross are inspected for the recessive phenotype. Gonadal streak
The appearance of the recessive phenotype identifies the (underdeveloped
chromosome that has one copy missing as the one the gonadal
gene is normally located on. The test works because half structures)
the gametes of a fertile monosomic will be n Ϫ 1, and
when an n Ϫ 1 gamete is fertilized by a gamete bearing No menstruation
a new mutation on the homologous chromosome, the
mutant allele will be the only allele of that gene present Figure 15-14 Characteristics of Turner syndrome.
and hence will be expressed. The condition results from the presence of a single X
chromosome (XO). [Adapted from F. Vogel and A. G. Motulsky,
As an example, let’s assume that a gene A/a is on
chromosome 2. Crosses of a/a to monosomics for Human Genetics. Springer-Verlag, 1982.]
chromosome 1 and chromosome 2 illustrate the method
(chromosome 1 is abbreviated chr1):

chr1/chr1 ; a/a ϫ chr1/0 ; A/A

Mutant Chromosome 1 monosomic us assume that the genotype is A/a/a. Furthermore, let’s
postulate that at anaphase I the two paired centromeres
genotype A in the trivalent pass to opposite poles and that the other
centromere passes randomly to either pole. Then we can
s predict the three equally frequent segregations shown in
p Figure 15-15. These segregations result in an overall ga-
metic ratio as shown in the six compartments of Figure
progeny all A/a 15-15; that is,

chr1/chr1 ; a/a ϫ chr1/chr1 ; A/0

Mutant Chromosome 2 monosomic

genotype A 1
6
s A

p 2
6
progeny 1 A/a a
2
2
1 a/0 6 A/a
2
1
6 a/a

TRISOMICS (2n ؉ 1) Trisomics contain an extra copy If a set of lines is available, each carrying a different
of one chromosome. In diploid organisms generally, the trisomic chromosome, then a gene mutation can be lo-
chromosomal imbalance from the trisomic condition can cated to a chromosome by determining which of the
result in abnormality or death. However, there are many lines gives a trisomic ratio of the above type.
examples of viable trisomics. Furthermore, trisomics can
be fertile. When cells from some trisomic organisms are There are several examples of viable human tri-
observed under the microscope at the time of meiotic somies. Several types of sex chromosome trisomics can
chromosome pairing, the trisomic chromosomes are live to adulthood. Each of these types is found at a fre-
seen to form an associated group of three (a trivalent), quency of about 1 in 1000 births of the relevant sex. (In
whereas the other chromosomes form regular pairs. considering human sex chromosome trisomies, recall
that mammalian sex is determined by the presence or
What genetic ratios might we expect for genes on absence of the Y chromosome.) The combination XXY
the trisomic chromosome? Let us consider a gene A that results in Klinefelter syndrome. Persons with this syn-
is close to the centromere on that chromosome, and let

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15.1 Changes in chromosome number 493

One pole Other pole behavior. Males with XYY are usually fertile. Meioses
show normal pairing of the X with one of the Y’s; the
A A a other Y does not pair and is not transmitted to gametes.
1 1 3 Therefore the gametes contain either X or Y, never YY
2a or XY. Triplo-X trisomics (XXX) are phenotypically
a a normal and fertile females. Meiosis shows pairing of only
2 A 2 two X chromosomes; the third does not pair. Hence eggs
1 bear only one X and, as in the case of XYY individuals,
a 3a A the condition is not passed on to progeny.
3 1
a Of human trisomies, the most familiar type is Down
2 syndrome (Figure 15-17), which we discussed briefly at
3a the beginning of the chapter. Down syndrome occurs at
a frequency of about 0.15 percent of all live births. Most
Figure 15-15 Genotypes of the meiotic products of an A/a/a affected individuals have an extra copy of chromosome
trisomic. The three segregations shown are equally likely. 21 caused by nondisjunction of chromosome 21 in a
parent who is chromosomally normal. In this sporadic
drome are males with lanky builds and a mildly im- type of Down syndrome, there is no family history of
paired IQ and are sterile (Figure 15-16). Another abnor- aneuploidy. Some rare types of Down syndrome arise
mal combination, XYY, has a controversial history. At- from translocations (a type of chromosomal rearrange-
tempts have been made to link the XYY condition with ment discussed later in the chapter); in these cases, as
a predisposition toward violence. However, it is now we shall see, Down syndrome recurs in the pedigree be-
clear that an XYY condition in no way guarantees such cause the translocation may be transmitted from parent
to child.
Tall stature Frontal baldness
Slightly feminized absent The combined phenotypes that make up Down syn-
physique Poor beard growth drome include mental retardation (with an IQ in the 20-
Mildly impaired IQ to-50 range); a broad, flat face; eyes with an epicanthic
(15 points less Breast development fold; short stature; short hands with a crease across the
than average) (in 30% of cases) middle; and a large, wrinkled tongue. Females may be
Tendency to lose Osteoporosis fertile and may produce normal or trisomic progeny, but
chest hairs males do not reproduce. Mean life expectancy is about
Small testes 17 years, and only 8 percent of persons with Down syn-
Female-type drome survive past age 40.
pubic hair
pattern Growth failure Broad flat face
Mental retardation Slanting eyes
Epicanthic eyefold
Flat back of head Short nose
Abnormal ears
Short and
Many "loops" broad hands
on fingertips Small and
Palm crease arched palate
Big, wrinkled
Special skin tongue
ridge patterns Dental anomalies
Unilateral or bilateral
absence of one rib Congenital heart
disease
Intestinal blockage
Enlarged colon
Umbilical hernia
Abnormal pelvis Big toes widely
spaced
Diminished muscle tone

Figure 15-16 Characteristics of Klinefelter syndrome (XXY). Figure 15-17 Characteristics of Down syndrome (trisomy 21).

[Adapted from F. Vogel and A. G. Motulsky, Human Genetics. [Adapted from F. Vogel and A. G. Motulsky, Human Genetics.
Springer-Verlag, 1982.] Springer-Verlag, 1982.]

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494 Chapter 15 • Large-Scale Chromosomal Changes

Incidence of Down syndrome per number of births 1/46 “faunlike” ears, a small jaw, a narrow pelvis, and rocker-
bottom feet; almost all babies with trisomy 18 die
1/100 within the first few weeks after birth. All other trisomics
die in utero.
1/290
1/2300 1/1600 1/1200 1/880 The concept of gene balance

0 20 25 30 35 40 45 In considering aberrant euploidy, we noted that an in-
Age of mother (years) crease in the number of full chromosome sets correlates
with increased organism size, but that the general shape
Figure 15-18 Maternal age and the production of offspring and proportions of the organism remain very much the
with Down syndrome. [From L. S. Penrose and G. F. Smith, Down’s same. In contrast, autosomal aneuploidy typically alters
the organism’s shape and proportions in characteristic
Anomaly. Little, Brown and Company, 1966.] ways.

The incidence of Down syndrome is related to ma- Plants tend to be somewhat more tolerant of aneu-
ternal age; older mothers run a greatly elevated risk of ploidy than are animals. Studies in jimsonweed (Datura
having a child with Down syndrome (Figure 15-18). For stramonium) provide a classic example of the effects of
this reason, fetal chromosome analysis (by amniocentesis aneuploidy and polyploidy. In jimsonweed, the haploid
or by chorionic villus sampling) is now recommended chromosome number is 12. As expected, the polyploid
for older mothers. A less pronounced paternal-age effect jimsonweed is proportioned like the normal diploid,
also has been demonstrated. only larger. In contrast, each of the 12 possible trisomics
is disproportionate, but in ways different from one an-
Even though the maternal-age effect has been other, as exemplified by changes in the shape of the seed
known for many years, its cause is still not known. capsule (see Figure 3-3). The 12 different trisomies lead
Nonetheless, there are some interesting biological corre- to 12 different and characteristic shape changes in the
lations. It is possible that with age it becomes less likely capsule. Indeed, these and other characteristics of the in-
that the chromosome tetrad will keep together during dividual trisomics are so reliable that the phenotypic
prophase I of meiosis. Meiotic arrest of oocytes (female syndrome can be used to identify plants carrying a par-
meiocytes) in late prophase I is a common phenomenon ticular trisomy. Similarly, the 12 monosomics are them-
in many animals. In female humans, all oocytes are ar- selves different from one another and from each of the
rested at diplotene before birth. Meiosis resumes at each trisomics. In general, a monosomic for a particular chro-
menstrual period, which means that the chromosomes mosome is more severely abnormal than is the corre-
in the tetrad must remain properly associated for as long sponding trisomic.
as several decades. If we speculate that these associations
have an increasing probability of breaking down by acci- We see similar trends in aneuploid animals. In the
dent over time, we can envision a mechanism contribut- fruit fly Drosophila, the only autosomal aneuploids that
ing to increased maternal nondisjunction with age. Con- survive to adulthood are trisomics and monosomics for
sistent with this speculation, most nondisjunction chromosome 4, which is the smallest Drosophila chromo-
related to the effect of maternal age is due to nondis- some, representing only about 1 to 2 percent of the
junction at anaphase I, not anaphase II. genome. Trisomics for chromosome 4 are only very
mildly affected and are much less abnormal than are
The only other human autosomal trisomics to sur- monosomics for chromosome 4. In humans, no auto-
vive to birth are those with trisomy 13 (Patau syn- somal monosomic survives to birth, but as we have seen,
drome) and trisomy 18 (Edwards syndrome). Both show three types of autosomal trisomics can do so. As is true of
severe physical and mental abnormalities. The pheno- aneuploid jimsonweed, these three trisomics each show
typic syndrome of trisomy 13 includes a harelip; a small, unique phenotypic syndromes because of the special ef-
malformed head; “rockerbottom” feet; and a mean life fects of altered dosages of each of these chromosomes.
expectancy of 130 days. That of trisomy 18 includes
Why are aneuploids so much more abnormal than
polyploids? Why does aneuploidy for each chromosome
have its own characteristic phenotypic effects? And why
are monosomics typically more severely affected than
are the corresponding trisomics? The answers seem cer-
tain to be a matter of gene balance. In a euploid, the
ratio of genes on any one chromosome to the genes on
other chromosomes always is 1 : 1, regardless of whether
we are considering a monoploid, diploid, triploid, or
tetraploid. For example, in a tetraploid, for gene A on

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15.1 Changes in chromosome number 495

chromosome 1 and gene B on chromosome 2, the ratio is alleles present on a monosomic autosome will be auto-
4 A : 4 B, or 1 : 1. In contrast, in an aneuploid, the ratio of matically expressed.
genes on the aneuploid chromosome to genes on the
other chromosomes differs from the wild type by 50 How do we apply the idea of gene balance to cases
percent; 50 percent for monosomics; 150 percent for tri- of sex chromosome aneuploidy? Gene balance holds for
somics. Using the same example as above, in a trisomic sex chromosomes as well, but we also have to take into
for chromosome 2, the ratio of the A and B genes is account the special properties of the sex chromosomes.
2A : 3B. Thus, we can see that the aneuploid genes are In organisms with XY sex determination, the Y chromo-
out of balance. How does this help us answer the ques- some seems to be a degenerate X chromosome in which
tions raised? there are very few functional genes other than some in-
volved in sex determination itself, in sperm production,
In general, the amount of transcript produced by a or in both. The X chromosome, on the other hand, con-
gene is directly proportional to the number of copies of tains many genes involved in basic cellular processes
that gene in a cell. That is, for a given gene, the rate of (“housekeeping genes”) that just happen to reside on
transcription is directly related to the number of DNA the chromosome that eventually evolved into the X
templates available. Thus, the more copies of the gene, chromosome. XY sex determination mechanisms have
the more transcripts are produced and the more of the probably evolved independently from 10 to 20 times in
corresponding protein product is made. This relationship different taxonomic groups. Thus, there appears to be
between the number of copies of a gene and the amount one sex determination mechanism for all mammals, but
of the gene’s product made is called a gene-dosage effect. it is completely different from the mechanism governing
XY sex determination in fruit flies.
We can infer that normal physiology in a cell de-
pends on the proper ratio of gene products in the eu- In a sense, X chromosomes are naturally aneuploid.
ploid cell. This ratio is the normal gene balance. If the In species with an XY sex determination system, females
relative dosage of certain genes changes — for example, have two X chromosomes, whereas males have only one.
because of the removal of one of the two copies of a Nonetheless, it has been found that the X chromosome’s
chromosome (or even a segment thereof ) — physiologi- housekeeping genes are expressed to approximately
cal imbalances in cellular pathways can arise. equal extents per cell in females and in males. In other
words, there is dosage compensation. How is this ac-
In some cases, the imbalances of aneuploidy result complished? The answer depends on the organism. In
from the effects of a few “major” genes whose dosage has fruit flies, the male’s X chromosome appears to be hy-
changed, rather than from changes in the dosage of all peractivated, allowing it to be transcribed at twice the
the genes on a chromosome. Such genes can be viewed rate of either X chromosome in the female. As a result,
as haplo-abnormal (resulting in an abnormal phenotype the XY male Drosophila has an X gene dosage equiva-
if present only once) or triplo-abnormal (resulting in an lent to that of an XX female. In mammals, in contrast,
abnormal phenotype if present in three copies) or both. the rule is that no matter how many X chromosomes
They contribute significantly to the aneuploid pheno- are present, there is only one transcriptionally active
typic syndromes. For example, the study of persons tri- X chromosome in each somatic cell. This rule gives
somic for only part of chromosome 21 has made it pos- the XX female mammal an X gene dosage equivalent
sible to localize genes contributing to Down syndrome to that of an XY male. Dosage compensation in mam-
to various regions of chromosome 21; the results hint mals is achieved by X-chromosome inactivation. A
that some aspects of the phenotype might be due to female with two X chromosomes, for example, is a
triplo-abnormality for single major genes in these chro- mosaic of two cell types in which one or the other
mosome regions. In addition to these major gene effects, X is active. We examined this phenomenon in Chapter
other aspects of aneuploid syndromes are likely to result 10. Thus, XY and XX individuals produce the same
from the cumulative effects of aneuploidy for numerous amounts of X-chromosome housekeeping-gene products.
genes whose products are all out of balance. Undoubt- X-chromosome inactivation also explains why triplo-X
edly, the entire aneuploid phenotype results from a humans are phenotypically normal — only one of the
combination of the imbalance effects of a few major three X chromosomes is transcriptionally active in a
genes, together with a cumulative imbalance of many given cell. Similarly, an XXY male is only moderately
minor genes. affected because only one of his two X chromosomes is
active in each cell.
However, the concept of gene balance does not tell
us why having too few gene products (monosomy) is Why are XXY individuals abnormal at all, given
much worse for an organism than having too many gene that triplo-X individuals are phenotypically normal? It
products (trisomy). In a parallel manner, we can ask why turns out that a few genes scattered throughout an “in-
there are many more haplo-abnormal genes than triplo- active X” are still transcriptionally active. In XXY males,
abnormal ones. A key to explaining the extreme abnor- these genes are transcribed at twice the level they are in
mality of monosomics is that any deleterious recessive